Answer:
Solution-
From the question the volume of water = 18 L = 18000 mL
Now we can find the mass of water = (volume of water) * (density of water)
mass of water = (18000 mL) * (1.00 g/mL)
mass of water = 18000 g
Now we find the heat required to boil water = (mass of water) * (specific heat water) * (final temperature - initial temperature)
putting the value the heat required to boil water = (18000 g) * (4.184 J/g.oC) * (100 oC - 22.7 oC)
heat required to boil water = 5821617.6 J
heat required to boil water = 5821.62 kJ
The heat given by the combustion = (heat required to boil water) / (percent of heat taken by boiling)
Heat given by combustion = (5821.62 kJ) / (19.4 /100)
Therefore the heat given by combustion = (5821.62 kJ) / (0.194)
Heat given by combustion = 30008.35 kJ
As we know that the enthalpy of combustion of methane = 802.5 kJ/mol
The moles methane used = (Heat given by combustion) / (enthalpy of combustion of methane)
moles methane used = (30008.35 kJ) / (802.5 kJ/mol)
So the moles methane used = 37.39 mol
Now the mass methane = (moles methane used) * (molar mass methane)
The mass methane = (37.39 mol) * (16.04 g/mol)
The mass methane = 599.74 g
Now the volume methane = (mass methane) / (density of methane)
volume methane = (599.7356 g) / (0.660 g/L)
volume methane = 908.69 L
hope helped!!
plz mark brainliest:DD
The volume of natural gas needed to boil the water if only 17.9% of the heat is generated towards heating water is ; 918.70 L
Using the given data :
Volume of water = 16.5 L = 16500 mL
mass of water = 16500 g ( 16500 mL * 1.00 g/mL )
Density of methane = 0.668 g/L
Density of water = 1.00 g/mL
First step : determine the heat needed to boil the water
Heat required = mass of water * specific heat water * ( Δ T )
= 16500 * 4.184 * ( 100 - 20.4 ) = 5495265.6 J
= 5495.265 kJ
∴ Heat required to boil water = 5495.265 kJ
next step ; determine the heat given by combustion
heat given by combustion = ( Heat required to boil water) / ( % of heat generated )
Heat given by combustion = ( 5495.265 ) / ( 17.9 % )
= 30699.80 kJ
Enthalpy of methane combustion = 802.5 kJ/mol
∴ moles of methane used = ( 30699.80 ) / ( 802.5 ) = 38.26 mol
next ; determine the mass of methane ( natural gas )
= ( moles of methane used ) * ( molar mass )
= 38.26 * 16.04 g/mol = 613.69 g
Final step : Calculate the volume of natural gas is needed to boil the water
= mass of natural gas / density of methane
= 613.69 g / 0.668 g/L
= 918.70 L
Hence we can conclude that the volume of natural gas needed to boil water if only 17.9% of the heat is 918.70 L .
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A compound containing only carbon, hydrogen, and oxygen is analyzed using combustion analysis. When 50.1 g of the compound is burned, 91.8 g of carbon dioxide and 25.1 g of water are collected. In order to determine the moles of carbon in the compound, first determine the moles of carbon dioxide that were produced from the combustion.
Answer:
[tex]C_{3}H_4O_2[/tex]
Explanation:
Hello,
In this case, since the carbon of the initial compound is present in the carbon dioxide product, we can compute the mass and moles of carbon in the compound:
[tex]n_C=91.8g CO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =2.09molC\\\\m_C=91.8g CO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2}*\frac{12gC}{1molC} =25.0gC[/tex]
Next, the mass and moles of hydrogen in the compound, is contained in the yielded amount of water, thus, we compute the mass and moles of hydrogen in the compound:
[tex]n_H=25.1gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =2.79molH\\\\m_H=25.1gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} *\frac{1gH}{1molH} =2.79gH[/tex]
In such a way, the mass of oxygen comes from the mass of the compound minus the mass of carbon and oxygen:
[tex]m_O=50.1g-25.0g-2.79g=22.31gO[/tex]
And the moles:
[tex]n_O=22.31gO*\frac{1molO}{16gO}=1.39molO[/tex]
Then, we compute the subscripts by diving the moles of C, H and O by the moles of oxygen as the smallest moles:
[tex]C:\frac{2.09}{1.39}=1.5 \\\\H:\frac{2.79}{1.39}=2\\ \\O:\frac{1.39}{1.39} =1[/tex]
After that, we write:
[tex]C_{1.5}H_2O[/tex]
Which must be shown in whole number only, thereby we multiply the subscripts by 2, so the empirical formula turns out:
[tex]C_{3}H_4O_2[/tex]
Best regards.
Answer each of the following questions with increases, decreases, or does not change.
a) If the intermolecular forces in a liquid increase, the normal boiling point of the liquid increases .
b) If the intermolecular forces in a liquid increase, the vapor pressure of the liquid does not change .
c) If the surface area of a liquid increases, the vapor pressure increases .
d) If the temperature of a liquid decreases, the equilibrium vapor pressure increases .
Drag and drop your selection from the following list to complete the answer:
does not change decreases increases
Answer:
a) increases
b) decreases
c) does not change
d) increases
Explanation:
The vapour pressure of a liquid is dependent on;
I) the magnitude of intermolecular forces
II) the temperature of the liquid
Hence, when any of these increases, the vapour pressure increases likewise.
Similarly, the boiling point of a liquid depends on the magnitude of intermolecular forces present because as intermolecular forces increases, more energy is required to break intermolecular bonds.
Lastly, increase in surface area of a liquid does not really affect it's vapour pressure.
A sample of an unknown gas weighs 0.419 grams and produced 5.00L of gas at 1.00atm (unknown gas only) and 298.15K, what is the molar mass (g/mole) of this unknown gas
Answer:
molar mass of unknown gas = 1.987 g/mol
Explanation:
First, the number of moles of the unknown gas is found
Using the ideal gas equation: PV = nRT
P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹
n = PV/RT
n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)
n = 0.2109 moles
Molar mass = mass/ number of moles
molar mass = 0.419 g/ 0.2109 mols
molar mass of unknown gas = 1.987 g/mol
The molar mass of unknown gas by using ideal gas equation = 1.987 g/mol.
Ideal gas equationThis equation gives the relation between pressure, volume, temperature as given below:
[tex]PV = nRT[/tex]
P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹
Substitute the above values in the above equation as follows:
n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)
n = 0.2109 moles
Formula for molar mass[tex]Molar mass = mass/ number of moles[/tex]
Calculate molar mass by using the above equation,
molar mass = 0.419 g/ 0.2109 moles
The molar mass of unknown gas = 1.987 g/mol
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What is the temperature of o3 gas if the average speed(actually the root mean square speed of the molecules is 750 m/s
Answer:
1.08 × 103 K
hope this helps :)
Explanation:
What is the pH of a bleach solution that has a [OH−]=1.3×10−4 M?
Answer:
pH = 10.113
Explanation:
Here, we can find pOH first:
pOH = [tex]-log([OH^-])[/tex] = 3.886.
Then, we can find pH which is 14 - pOH. We then get the answer above.
Then, we can find pH which is 14 - pOH. We then get the answer above.
The pH of acid is between 0-7 on pH scale while for base pH range is from 7-14. Thus the pH of 1.3×10⁻⁴ M bleach solution is 10.62.
What is pH?pH is a measurement of amount of hydronium ion H₃O⁺ in a given sample. More the value of hydronium ion concentration, more will be the solution acidic.
On subtracting pH from 14, we get pOH which measures the concentration of hydroxide ion in a given solution. pH depend on the temperature. At room temperature pH scale is between 0 to 14. pH of neutral solution is 7
The concentration of bleach solution is 1.3×10⁻⁴ M
Concentration of OH⁻=1.3×10⁻⁴ M
Mathematically,
pOH=-log[OH⁻]
Substituting the values
pH=-log[1.3×10⁻⁴]
= 3.886.
pH+ POH=14
pH=14-3.886.=10.62
Therefore, the pH of 1.3×10⁻⁴ M bleach solution is 10.62.
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ILL GIVE BRAINLIEST IF CORRECT!!! students who enjoy acting out plays likely think with what part of their brains?
A. Back Brain
B. Front Brain
C. Left Brain
D. Right Brain
Answer:
B
Explanation:
The front brain(cerebrum) is the part of the brain that controls important cognitive skills in humans, such as emotional expression, problem solving, memory, language, judgment, and sexual behaviors.
Here the students who enjoy acting out plays likely think with front brain. I guess so..
Answer: The front brain(cerebrum) is the part of the brain that controls important cognitive skills in humans, such as emotional expression, problem solving, memory, language, judgment, and sexual behaviors
Explanation: people who do acting or choir would use the front brain..... i think :) hope this helps.
Choose all the answers that apply.
Fluorine (F) has seven electrons in its outermost shell. Fluorine _____.
-is a metal
-fills its shell by giving up seven electrons
-fills its shell by gaining one electron
-becomes a positively charged ion
-becomes a negatively charged ion
-is a nonmetal
Answer:
Fluorine is a non-metal, fills its shell by gaining one electron and becomes a negatively charged ion.
Explanation:
Fluorine have seven electrons in its outermost shell. This means that fluorine is located in Group 17(Halogens) in the Periodic Table. Note that halogens are non-metal elements. So fluorine is a non-metal.
Fluorine fills its shell by gaining one electron because it only needs one electron to achieve stable octet electron arrangement. ( it's easier to gain one electron than to lose all seven electrons in the outermost shell)
As fluorine atom gain one electron to achieve stable octet electron arrangement, its number of electrons becomes greater than its number of protons. So it becomes a negatively charged ion.
Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.
This question is incomplete, the complete question is;
Tonksite is a solid at 300.00K. At 300.00 K its enthalpy of sublimation is 66.00 kJ/mol. The sublimation pressure at 300.00 K is 5.00 × 10⁻⁴ atm
Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.
Answer: the sublimation pressure of the solid at the melting point is 0.3727 atm
Explanation:
Given that;
T1 = 300 K
T2 = 400 K
H_sub = 66 kJ/mol = 66000 J/mol
P1 = 5.00 × 10⁻⁴ atm
p2 = ?
now using the expression
log( p2 / 5.00 × 10⁻⁴ ) = (H_sub / R × 2.303 ) (( T2 - T1) / T1T2)
now we substitute of given values into the expression
log(p2/p1) = (66000 / 8.314 × 2.303 ) (( 400 - 300) / 300 × 400 )
p2 = 0.3727 atm
therefore the sublimation pressure of the solid at the melting point is 0.3727 atm
What is the heat gjoules transferred by a chemical reaction to the reservoir of a calorimeter containing 135g of dilute aqueous solution (c=4.184 J/g^ C) the reaction causes the temperature of the reservoir to rise from 23.0 27.0 degrees * C
Answer:
Required heat = 2,259.36 N (approx)
Explanation:
Given:
Mass = 135 g
Specific heat (c) =4.184 J/gC
Change in temperature ΔT = 27-23 = 4 c
Find:
Required heat
Computation:
Q = mcΔT
Q = (135)(4.184)(4)
Required heat = 2,259.36 N (approx)
The heat transferred by the chemical reaction to the reservoir of a calorimeter containing 135 g of dilute aqueous solution is 2259.36 J
To solve this question, we'll begin by calculating the change in the temperature of the reservoir. This can be obtained as follow:
Initial temperature (T₁) = 23 °C
Final temperature (T₂) = 27 °C
Change in temperature (ΔT) =?ΔT = T₂ – T₁
ΔT = 27 – 23
ΔT = 4 °CFinally, we shall determine the heat transferred to the reservoir.
Change in temperature (ΔT) = 4 °C
Mass (M) = 135 g
Specific heat capacity (C) = 4.184 J/gºC
Heat (Q) =?Q = MCΔT
Q = 135 × 4.184 × 4
Q = 2259.36 JTherefore, the heat transferred by the chemical reaction to the reservoir of a calorimeter containing 135 g of dilute aqueous solution is 2259.36 J
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HELPPPPP
Identify whether each of the following changes is a physical change or a chemical change. Write “P” on the line for a physical change and a “C” for a chemical change.
22. Water boiling ______23. Iron rusting ______24. Butter melting _____25. Wood rotting ______
26.Alcohol evaporating _____27.Glass breaking ______28.Mowing the lawn ______29.Baking a cake ______
Answer:
22 is P
23 is C
24 is P
25 is C
26 is P
27 is P
28 is P
29 is C because youre using thermal heat and you cannot return to raw batter
C. Where is sugar removed from the blood?
and how can you tell
Answer:During absorption and digestion, the carbohydrates in the food you eat are reduced to their simplest form, glucose. Excess glucose is then removed from the blood, with the majority of it being converted into glycoge, the storage form of glucose, by the liver's hepatic cells via a process called glycogenesis
Explanation:
Help plzzzzzzzzzzzzzzz
Answer:
um wheres thepassage?
Explanation:
The heat of fusion for naphthalene is 0.148 kJ/g. Calculate the heat energy in joules required to melt 52.95 g of naphthalene.
Answer:
7836.6 J
Explanation:
The following data were obtained from the question:
Heat of fusion (Hf) = 0.148 KJ/g
Mass (m) = 52.95 g
Heat (Q) required =..?
The heat (Q) required to melt the sample of nephthalene can be obtained as follow:
Q = m·Hf
Q = 52.95 × 0.148
Q = 7.8366 KJ
Finally, we shall convert 7.8366 KJ to Joule (J) in order to obtain the desired result. This can be obtained as follow:
1 kJ = 1000 J
Therefore,
7.8366 KJ = 7.8366 KJ × 1000 J / 1 KJ
7.8366 KJ = 7836.6 J
Therefore, 7.8366 KJ is equivalent to 7836.6 J
Thus, 7836.6 J of heat energy is required to melt the sample of nephthalene.
The heat required for melting of 52.95 g of naphthalene has been 7.8366 kJ.
The heat of fusion has been the energy required by 1g for the change in the substance.
The heat required to melt naphthalene can be given as:
Heat = mass × Heat of fusion
The given mass = 52.95 g
The heat of fusion = 0.148 kJ/g
The heat required can be given as:
Heat = 0.148 × 52.95 kJ
Heat = 7.8366 kJ.
The heat required for melting of 52.95 g of naphthalene has been 7.8366 kJ.
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Determine the mass (in grams) of NaCl in 294 grams of a 24.1% (m/m) NaCl solution. Be sure to report to the correct number of significant figures with no units.
I NEED HELP WITH THIS URGENTLY!!!!!
Answer:
70.9 grams 3 sig figs
Explanation:
24.1% of 294 grams = 0.241(294 gram) = 70.854 grams ≅ 70.9 grams 3 sig figs
What is magma?
solid rock with a fine texture
cooled rock with large crystals
molten rock below Earth’s surface
flowing rock above Earth’s surface
Answer:
C. molten rock below Earth’s surface
Explanation:
Which is NOT an example of plants demonstrating the characteristics of life?
A. Plants are forces to make adaptations depending on its environment
B. A plant has stimuli that cause it to grow towards the sun
C. There are stages of growth that plants go through depending on favorable conditions
D. The leaves on a plant move in reaction to the wind
Answer:
D
Explanation:
trust me bro ive done this
Option D does not represent an example of plants demonstrating the characteristics of life.
The following information should be considered:
Plants to be treated as the forces for making the adaptions and based on the environment. The plant contains stimuli that result to grow towards the sun. There should be the growth stages based on the favorable conditions.Therefore we can conclude that Option D does not represent an example of plants demonstrating the characteristics of life.
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1. What is the problem in this activity?
2. Formulate your hypothesis. (List down at least 3 hypotheses)
3. What is the dependent variable?
4. What is the independent variable?
5. Analyze your results. Which type of liquid will prevent the apple from
turning brown?
a. lemon
b. milk
Hello. You forgot to say that this question is about the possibility of using some type of product that prevents cut apples from becoming darker.
Answer and Explanation:
1. The problem with this activity is "What can you put on an apple slice to keep it from turning brown?" The problem in a scientific experiment is the element that provides a question about what is being observed.
2. Hypotheses are assumptions made about what was observed, in addition, hypotheses can answer the question shown in the problem. In the case of apples, the hypotheses can be: "Lemon juice is the most efficient product in delaying the blackening of cut apples." "Water is inefficient in preventing blackened cut apples." "Milk accelerates the blackness of cut apples."
3. The dependent viable is one that needs to be influenced by another element to bring about a result. In the case of the experiment with apples, the dependent variable is the blackening rate of the apple slices.
4. The independent variable, on the other hand, refers to the element that does not need any influence to cause a result, on the contrary, this variable acts on the dependent variable, influencing it. In the case of this experiment, the independent variable is the types of liquid tested to delay the blackening of cut apples.
5. Lemon was the most efficient liquid. Probably this efficiency is related to the lemon's ability to decrease the strength of antioxidants, due to its very low pH.
noble gas notation for titanium
Answer:
Titanium [Ar] 3d2 4s2
Explanation:
A 13.0-L scuba diving tank contains a helium-oxygen (heliox) mixture made up of 23.6 g of He and 4.85 g of O2 at 298 K. Calculate the mole fraction of each component in the mixture.
Answer:
[tex]x_{He}=0.975\\x_{O_2}=0.025[/tex]
Explanation:
Hello.
In this case, since we know the mass of both helium and oxygen, in order to obtain the mole fractions we first need the compute the moles by using their atomic masses, 4.00 g/mol and 32.00 g/mol respectively as shown below:
[tex]n_{He}=23.6gHe*\frac{1molHe}{4.00gHe}=5.90molHe\\ \\n_{O_2}=4.85gO_2*\frac{1molO_2}{32.00gO_2}=0.152molO_2\\[/tex]
Therefore, the mole fractions are:
[tex]x_{He}=\frac{n_{He}}{n_{He}+n_{O_2}}=\frac{5.90}{5.90+0.152} \\\\x_{He}=0.975\\\\x_{O_2}=\frac{n_{O_2}}{n_{He}+n_{O_2}} =\frac{0.152}{5.90+0.152} \\\\x_{O_2}=0.025[/tex]
Best regards!
) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H5OH? What is the difference in pH values for the two acids?
Answer:
1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).
2. The difference in pH values is 4.95.
Explanation:
1. The pH of a compound can be found using the following equation:
[tex] pH = -log([H_{3}O^{+}]) [/tex]
First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.
Trimethyl ammonium:
We can calculate [H₃O⁺] using the Ka as follows:
(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺
1.0 - x x x
[tex]Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}[/tex]
[tex] 10^{-pKa} = \frac{x*x}{1.0 - x} [/tex]
[tex] 10^{-9.80}(1.0 - x) - x^{2} = 0 [/tex]
By solving the above equation for x we have:
x = 0.097 = [H₃O⁺]
[tex] pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01 [/tex]
Phenol:
C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺
1.0 - x x x
[tex]Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}[/tex]
[tex] 10^{-10} = \frac{x^{2}}{1.0 - x} [/tex]
[tex] 1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0 [/tex]
Solving the above equation for x we have:
x = 9.96x10⁻⁶ = [H₃O⁺]
[tex] pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00 [/tex]
Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.
2. The difference in pH values for the two acids is:
[tex] \Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95 [/tex]
Therefore, the difference in pH values is 4.95.
I hope it helps you!
Write the letter of the correct term for each definition in the blank. Use each term only once.
a. filtration
b. distillation
c. decantation
d. evaporation
e. centrifugation
f. chromatography
1. Separates substances on the basis of the boiling points of the substances:__________
2. Separates solutions based on boiling points when one substance remains a solid:__________
3. Separates pure substances from a mixture based upon different rates of flow:__________
4. Separates solids from liquids by pouring the liquid off the top of the solid:__________
5. Separates solids from liquids by using a porous barrier:__________
6. Separates heterogeneous substances based upon different densities:__________
Answer:
1.b. distillation
2.d. evaporation
3.f. chromatography
4.c. decantation
5.a. filtration
6.e. centrifugation
Explanation:
1. Distillation is the process in which a homogenous mixture of liquid is separated on the basis of difference in boiling points of the substances.
2. Evaporation is the process in which water is converted into gas or vapor based on the boiling point of water or solid.
3. Chromatography is defined as the process used to pure substances from a mixture based on the flow rate of the fluid. The fluid mixture to be separated is called the mobile phase that flows on a stationary phase.
4. Decantation is defined as the process that separates mixtures in which water is poured off the substance to be separated or over solid.
5. Filtration is a separation process in which solids can be separated from liquids by pouring the mixture over a porous barrier.
6. Centrifugation is defined as the separation process that separates heterogeneous substances on the basis of densities in which centripetal acceleration allows substance with high density to move outward in the radial direction.
32. A chemical reaction in which calcium carbonate CaCO3 is decomposed, results in the
production of two simpler compounds.
CaCO3
>CaO
+
CO2
? 9
28.00 g
21.97 g
What mass of calcium carbonate, to the nearest hundredth of a gram, is
decomposed in this reaction?
Answer:
Mass of calcium carbonate decompose 49g.
Explanation:
Given data:
Mass of CaO = 28.00 g
Mass of CO₂ = 21.97 g
Mass of calcium carbonate decompose = ?
Solution:
Chemical equation:
CaCO₃ → CO₂ + CaO
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 21.97 g / 44 g/mol
Number of moles = 0.49 mol
Number of moles of CaO:
Number of moles = mass/molar mass
Number of moles = 28.00 g / 56.1 g/mol
Number of moles = 0.49 mol
Now we will compare the moles of CaCO₃ with CaO and CO₂ .
CO₂ : CaCO₃
1 : 1
0.49 : 0.49
CaO : CaCO₃
1 : 1
0.49 : 0.49
Mass of CaCO₃ decompose:
Mass = number of moles × molaer mass
Mass = 0.49 mol × 100.1 g/mol
Mass = 49 g
Answer:
CaO
Explanation:
the coefficient of 02 is needed to balance the equation C3h8+O2+3co2+4H20
Answer:The anwser is 5
Explanation:
If an object has a density of 6.05 g/cm3 and a volume of 36.5 cm3, what is its mass?
Answer:
0.42 g/cm3
Explanation:
Answer:
20/40=0.5 g/cm^3 becuase, mass/volume=density
Explanation:
Vocabulary Terms: Please copy down the definitions for words below from the Smart Screen.
Length is defined as
Height is defined as
Circumference is defined as
Perimeter is defined as
Diagonal is defined as
Diameter is defined as
Answer:
Explanation:
Length: Length is a measure of distance
Height:The measurement from base to top or (of a standing person) from head to foot.
Circumference:The enclosing boundary of a curved geometric figure, especially a circle
Perimeter: Length of the outline of a shape.
Diagonal:A line segment that goes from one corner to another, but is not an edge. So when we directly join any two corners (called "vertices") which are not already joined by an edge, we get a diagonal. Diagonals of Polygons.
Diameter: A straight line passing from side to side of any figure or body, through its center.
Which of the following best support the function of electrical signals in the nervous system?
Each nerve cell is connected to the brain by electrical signals.
Nerve cells communicate with the brain by sending out electrical signals
Electrical signals keep nerve cells alive.
Electrical signals are used to get rid of waste produced in the nervous system.
Answer:
Each nerve cell is connected to the brain by electrical signals
Explanation:
Answer:
they said it above me
Explanation:
Explain How does its structure affect its function
Answer:
What are we talking about here?
Explanation:
1.547 grams of hydrated MgSO4 is heated in a crucible. After heating, 0.7554 grams of anhydrous MgSO4 remains in the crucible. How many waters of hydration were attached to the MgSO4
Answer:
7
Explanation:
Let x represent the number of moles of water in the hydrated salt i.e MgSO₄.xH₂O
The following data were obtained from the question:
Mass of MgSO₄.xH₂O = 1.547 g
Mass of anhydrous MgSO₄ = 0.7554 g
Mole of H₂O = x =?
Next, we shall determine the mass of water, H₂O in the hydrated salt, MgSO₄.xH₂O. This can be obtained as follow:
Mass of MgSO₄.xH₂O = 1.547 g
Mass of anhydrous MgSO₄ = 0.7554 g
Mass of H₂O =?
Mass of H₂O = (Mass of MgSO₄.xH₂O) – (Mass of anhydrous MgSO₄)
Mass of H₂O = 1.547 – 0.7554
Mass of H₂O = 0.7916 g
Finally, we shall determine the value of the x as illustrated below:
Mass of MgSO₄.xH₂O = 1.547 g
Molar mass of MgSO₄.xH₂O = 24 + 32 + (16×4) + x[(2×1) + 16]
= 24 + 32 + 64 + x(2 + 16)
= 120 + 18x
Mass of H₂O = 0.7916 g
Molar mass of xH₂O = 18x
Molar Mass of xH₂O/ Molar mass of MgSO₄.xH₂O = mass of xH₂O /Mass of MgSO₄.xH₂O
18x/ 120 + 18x = 0.7916/1.547
Cross multiply
0.7916 (120 + 18x) = 18x × 1.547
94.992 + 14.2488x = 27.846x
Collect like terms
94.992 = 27.846x – 14.2488x
94.992 = 13.5972x
Divide both side by 13.5972
x = 94.992 / 13.5972
x = 7
Thus, the formula for the hydrated salt, MgSO₄.xH₂O is MgSO₄.7H₂O
Number of moles of water, H₂O in the hydrated salt MgSO₄.7H₂O is 7.
The number of moles of attached water molecules is 7.
Mass of hydrated MgSO4 = 1.547 grams
Mass of anhydrous MgSO4 = 0.7554 grams
Number of moles of hydrated MgSO4 = 1.547 grams/120 + 18x
Number of moles of anhydrous MgSO4 = 0.7554 grams /120
Number of moles of anhydrous salt = Number of moles of hydrated salt
0.7554 grams /120 = 1.547 grams/120 + 18x
0.7554(120 + 18x) = 1.547 × 120
90.6 + 13.6x = 185.6
185.6 - 90.6 /13.6 = x
x = 7
The number of moles of attached water molecules is 7.
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Element
Molar mass (g/mol)
1.008
H
С
12.01
16.00
Using the information in the table, calculate the number of moles in a 2.03 kg sample of citric acid
(C6H8O7)
Write your answer using three significant figures.
mol C6H8O7-
Answer:
Number of moles = 10.6 mol
Explanation:
Given data:
Molar mass of H = 1.008 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Mass of citric acid = 2.03 kg (2.03×1000 = 2030 g)
Number of moles of citric acid = ?
Solution:
Formula:
Number of moles = mass/molar mass
Now we will calculate the molar mass of citric acid:
C₆H₈O₇ = (12.01× 6) + (1.008×8) + (16.00×7)
C₆H₈O₇ = 72.06 + 8.064+112
C₆H₈O₇ = 192.124g/mol
Number of moles = 2030 g/ 192.124g/mol
Number of moles = 10.6 mol
The number of moles in a 2.03 kg sample of citric acid is :
- 10.6 mol
Molar MassGiven:
Molar mass of H = 1.008 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Mass of citric acid = 2.03 kg (2.03×1000 = 2030 g)
Number of moles of citric acid = ?
Formula:
Number of moles = mass/molar mass
The molar mass of citric acid:
C₆H₈O₇ = (12.01× 6) + (1.008×8) + (16.00×7)
C₆H₈O₇ = 72.06 + 8.064+112
C₆H₈O₇ = 192.124g/mol
Number of moles = 2030 g/ 192.124g/mol
Number of moles = 10.6 mol
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A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 116.9 CWhat is the specific heat (SH) of the substance?
Answer:
Approximately [tex]0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}[/tex].
Explanation:
The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)
Calculate the increase in the temperature of this sample:
[tex]\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C[/tex].
The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)
It took [tex]2911\; \rm J[/tex] of energy to raise the temperature of this sample by [tex]\Delta T = 92.6\; \rm ^\circ\! C[/tex]. Therefore, raising the temperature of this sample by [tex]1\; \rm ^\circ\! C[/tex] (unit temperature) would take only [tex]\displaystyle \frac{1}{92.6}[/tex] as much energy. That corresponds to approximately [tex]31.436\; \rm J[/tex] of energy.
On the other hand, the energy required to raise the temperature of this material by [tex]1\; \rm ^\circ\! C[/tex] is proportional to the mass of the sample (also assuming no phase change.)
It took approximately [tex]31.436\; \rm J[/tex] of energy to raise the temperature of [tex]57.07\; \rm g[/tex] of this material by [tex]1\; \rm ^\circ C[/tex]. Therefore, it would take only [tex]\displaystyle \frac{1}{57.07}[/tex] as much energy to raise the temperature of [tex]1\; \rm g[/tex] (unit mass) of this material by [tex]1\; \rm ^\circ \! C\![/tex]. That corresponds to approximately [tex]0.551\; \rm J[/tex] of energy.
In other words, it takes approximately [tex]0.551\; \rm J[/tex] to raise [tex]1\; \rm g[/tex] (unit mass) of this material by [tex]1\; \rm ^\circ \! C[/tex]. Therefore, by definition, the specific heat of this material would be approximately [tex]0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}[/tex].