Answer:
10 moles of water are produced,
Explanation:
Given data:
Moles of H = 10 mol
Moles of O = 6 mol
Water obtained = ?
Solution:
Balance chemical equation:
2H₂ + O₂ → 2H₂O
Now we will compare the moles of H₂ and O₂ with water from balance chemical equation.
H₂ : H₂O
2 : 2
10 : 10
O₂ : H₂O
1 : 2
6 : 2×6 = 12
Number of moles of water produced by hydrogen are less so hydrogen will be limiting reactant and it will limit the yield of water thus, 10 moles of water are produced.
How do weathering and deposition differ? (4 points)
оа
Weathering breaks down rocks; deposition leaves them in new places.
Ob
Weathering has to do with air; deposition has to do with plants.
Ос
Weathering occurs only in summer; deposition occurs year-round.
Od
Weathering can be chemical or physical; deposition is only chemical.
Answer:
Oa. Weathering breaks down rocks; deposition leaves them in new places.
Explanation:
Did test and got it right.
Discuss the relationship between atoms, elements and compounds. Include in your discussion if these are mixtures or pure substances and why.
Answer:
Elements are the simplest complete chemical substances. Each element corresponds to a single entry on the periodic table. An element is a material that consists of a single type of atom. Each atom type contains the same number of protons.
Chemical bonds link elements together to form more complex molecules called compounds. A compound consists of two or more types of elements held together by covalent or ionic bonds.
Explanation:
What element in the second period has the largest atomic radius?
a
neon
b
lithium
c
carbon
d
potassium
Answer:
b) Lithium
Explanation:
I need help plzzz this is for today please help
Answer:
4. Water from upland areas often carries sediment and pollutants. The marshy land and plants in estuaries filter these pollutants out of the water. The plants in estuaries help prevent shoreline erosion. Estuaries also protect inland areas from flooding and storm surges.
5. Dredging impacts marine organisms negatively through entrainment, habitat degradation, noise, remobilization of contaminants, sedimentation, and increases in suspended sediment concentrations.
Explanation:
The farther you go in to the ocean, the more salinity it contains. The closer you are to freshwater, the less salinity it contains. This can happen in estuaries, and estuaries are almost like brackish waters.
Hope this helps you!
Which of the following molecules may show a pure rotational microwave absorption spectrum: (i) H2, (ii) HCl, (iii) CH4, (iv) CH3Cl, (v) CH2Cl2?
Answer:
Explanation:
H₂ , HCl , CH₃Cl , CH₂Cl₂ ---- Yes
CH₄ ------ No .
Symmetric molecules do not have rotational microwave absorption spectrum because they do not possess permanent dipole moment so they lack polarisability .
The pure rotational microwave absorption spectrum has been possessed by HCl and [tex]\rm \bold{CH_3Cl}[/tex].
Pure rotational microwave absorption has been consisted of the molecules having dipole and polarizability. The molecules that show the spectrum and dipole are diatomic molecules, linear molecules, spherical molecules.
(i) The hydrogen has been a diatomic molecule with symmetry thus not showing a pure rotational microwave absorption spectrum.
(ii) The HCl being a linear molecule possesses a pure rotational microwave absorption spectrum.
(iii) [tex]\rm CH_4[/tex] molecule has symmetry, thus will not possess a pure rotational microwave absorption spectrum.
(iv) [tex]\rm CH_3Cl[/tex] molecule has no symmetry, and thus possesses a pure rotational microwave absorption spectrum.
(v) [tex]\rm CH_2Cl_2[/tex] molecule with the presence of symmetry, thus lacks a pure rotational microwave absorption spectrum.
For more information about the pure rotational microwave absorption spectrum, refer to the link:
https://brainly.com/question/14215380
A solution has a pH of 6. What is true about the solution?
A. It is a strong basic solution.
B. It is a weak acidic solution.
C. It is a weak basic solution.
D. It is a strong acidic solution.
please help me
Answer:
A. it is a strong basic solution
Answer:
(see below)
Explanation:
First, refer to the pH scale:
1 2 3 4 5 6 7 8 9 10 11 12 13 14
<== acidic neutral basic ==>
You can see that the smaller the number, the stronger the acid and the bigger the number, the more basic the base is. 7 is neutral, such as water. it's neither basic nor acidic.
Now, using the process of elimination:
A) It's a strong basic solution.
No, because this solution's pH hasn't even reached basic.
B) It's a weak acidic solution.
Yes, because it is acidic and it's just a little bit more acidic than a neutral solution.
C) It's a weak basic solution.
No, because this solution's pH hasn't even reached basic.
D) It's a strong acidic solution.
No, because even though it's acidic, it's just below neutral. For something to be a strong acidic solution would be around a pH of 3.
So the answer would be B) It's a weak acidic solution.
List the following bonds in order of increasing ionic character: potassium to iodine, carbon to oxygen, lithium to fluorine, boron to fluorine. (Enter the two elements of the bond into the appropriate box: KI, CO, LiF, BF)
Answer:
CO < BF < KI < LiF
Explanation:
The magnitude of ionic character in a bond is dependent on the magnitude of electronegativity difference between the atoms in the bond.
Remember that no bond is 100% ionic or covalent according to Linus Pauling. However, the percentage ionic character depends on electronegativity difference between the bonding atoms and polarizability (Fajan's rules).
Between LiF and KI, Fajan's rules become very important. The Li^+ is small and highly polarizing. The stronger the polarising power of the cation and the higher the polarisability of the anion the more covalent character is expected in a bond
for the following reaction, provide the missing information
Answer:
19. Option B. ⁰₋₁B
20. Option D. ²¹⁰₈₄Po
Explanation:
19. ²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ
Thus, we can determine ʸₓZ as follow:
228 = 228 + y
Collect like terms
228 – 228 = y
y = 0
88 = 89 + x
Collect like terms
88 – 89 = x
x = –1
Thus,
ʸ ₓZ => ⁰₋₁Z => ⁰₋₁B
²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ
²²⁸₈₈Ra —> ²²⁸₈₉Ac + ⁰₋₁B
20. ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He
Thus, we can determine ᵘᵥX as follow:
u = 206 + 4
u = 210
v = 82 + 2
v = 84
Thus,
ᵘᵥX => ²¹⁰₈₄X => ²¹⁰₈₄Po
ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He
²¹⁰₈₄Po —> ²⁰⁶₈₂Pb + ⁴₂He
A 7.27-gram sample of a compound is dissolved in 250. grams of benzene. The freezing point of this solution is 1.02°C below that of pure benzene. What is the molar mass of this compound? (Note: Kf for benzene = 5.12°C/m.) Ignore significant figures for this problem. Group of answer choices 36.5 g/mol 146 g/mol 292 g/mol 5.79 g/mol 73.0 g/mol
Answer:
The correct answer is 146 g/mol
Explanation:
Freezing point depression is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:
ΔTf = Kf x m
Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:
ΔTf = 1.02ºC
Kf = 5.12ºC/m
From this, we can calculate the molality:
m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m
The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:
0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute
There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:
molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol
Therefore, the molar mass of the compound is 146 g/mol
In the combustion of hydrogen gas, hydrogen reacts with oxygen from the air to form water vapor. hydrogen+oxygen⟶water If you burn 58.9 g of hydrogen and produce 526 g of water, how much oxygen reacted?
Answer:
[tex]m_{O_2}=467gO_2[/tex]
Explanation:
Hello.
In this case, for the reaction:
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
The correct way to compute the oxygen that reacted is by considering the mass of hydrogen, 58.9 g, its molar mass, 2.02 g/mol, the 2:1 mole ratio between hydrogen and oxygen and the atomic mass of that gaseous oxygen 32.00 g/mol; therefore we use the following stoichiometric procedure:
[tex]m_{O_2}=58.9gH_2*\frac{1molH_2}{2.02gH_2}*\frac{1molO_2}{2molH_2} *\frac{32.00gO_2}{1molO_2} \\\\m_{O_2}=467gO_2[/tex]
The same result could have been obtained by using the mass of water since the law of conservation of mass is obeyed here.
Best regards!
what are molecular compounds
Answer:
inorganic compounds that take the form of discrete molecules
Answer:
Compounds that take the form of discrete molecules.
Explanation:
Rather than forming ions, the atoms of a molecule share their electrons in such a way that a bond forms between pairs of atoms.
A volume of 80.0 mL of a 0.690 M HNO3 solution is titrated with 0.790 M KOH. Calculate the volume of KOH required to reach the equivalence point. Express your answer to three significant figures and include the appropriate units.
Given :
A volume of 80.0 mL of a 0.690 M [tex]HNO_3[/tex] solution is titrated with 0.790 M KOH.
To Find :
The volume of KOH required to reach the equivalence point.
Solution :
We know, at equivalent point :
moles of [tex]HNO_3[/tex] = moles of KOH
[tex]M_{HNO_3}V_{HNO_3}=M_{KOH}V_{KOH}\\\\0.690\times 80 = 0.790\times V_{KOH}\\\\V_{KOH}=\dfrac{0.690\times 80 }{ 0.790}\ ml\\\\V_{KOH}=69.87\ ml[/tex]
Therefore, volume of KOH required is 69.87 ml.
Hence, this is the required solution.
an oxide of copper is decomposed forming copper metal and oxygen gas. a 0.500 g sample of this oxide is decomposed, forming 0.444 g of copper metal. what is the empircal formula of the gold oxide
Answer:
[tex]Cu_2O[/tex]
Explanation:
Hello.
In this case, since the reaction obeys the law of conservation of mass, since the initial oxide had a mass of 0.500 g and yielded 0.444 g of copper, the mass of oxygen in the oxide is:
[tex]m_O=0.500g-0.444g=0.056g[/tex]
In such a way, we next compute the moles of copper and oxygen by using their atomic masses:
[tex]n_{Cu}=0.444g*\frac{1mol}{63.546 g} =0.00699mol\\\\n_O=0.056g*\frac{1mol}{16.00g}=0.0035mol[/tex]
Next, in order to compute the subscripts of Cu and O on the empirical formula we divide the moles by the fewest moles, in this 0.0035 mol as shown below:
[tex]Cu:\frac{0.00699}{0.0035} =2\\\\O:\frac{0.0035}{0.0035} =1[/tex]
It means that the empirical formula turns out:
[tex]Cu_2O[/tex]
Best regards!
which dissolved first in acetone? food coloring or liquid paint?
Answer:
Liquid paint.
Explanation:
Liquid paints are dissolved first in acetone. Most of the food dyes are not soluble in acetone.
What is acetone?Acetone is an organic compound comes under the category of ketones. It contains a carbonyl group and can dissolve most of the organic solvents.
The active ends in acetone easily forms hydrogen bonds with other solvents with polar or nonpolar groups.
Xylene, benzene, toluene, aromatic azo dyes etc. are common components in paint. Which are easily miscible with acetone.
Hence, liquid paints dissolve in acetone and food dyes are hard o dissolve in it.
To learn more about acetone, refer the link below:
https://brainly.com/question/13334667
#SPJ2
which element has the highest ionization energy?B, Al, Ga, In
Answer:
BORON
Ionization energy decreases down the group and going from left to right the period,
Luckily you have the elements from the same group that is Group III A also called boron family,
The position of Elements in this group are
Boron (B)Aluminium (Al)Gallium (Ga)Indium(In)ThalliumNihoiumso keeping rules in mind the first element in the group has highest I.E. that is boron
What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl
Answer:
70.88 mL volume of 1.27 M of HCl is required.
Explanation:
Given data:
Initial volume = ?
Initial molarity = 1.27 M
Final volume = 197.4 mL
Final molarity = 0.456 M
Solution:
Formula:
M₁V₁ = M₂V₂
Now we will put the values in formula.
1.27 M × V₁ = 0.456 M × 197.4 mL
V₁ = 0.456 M × 197.4 mL/1.27 M
V₁ = 90.014M.mL/1.27 M
V₁ = 70.88 mL
70.88 mL volume of 1.27 M of HCl is required.
Use the image above to answer the
question.
1. (6B) What two particles are found in the
nucleus of an atom? (choose two options)
Protons
Neutrons
Nucleus
Electron
Electron cloud
Answer:
A-protons, B-nucleus, D-electrons, C-nucleus, E-electric cloud
Explanation:
can someone pweaseee help me on this ??
Answer:
Be
Explanation:
The atomic radius decreases as you go from left to right in a period. Be has the largest radius out of those given elements.
Which element is more reactive?
A) Flourine B) Oxygen C)Cabron D)Boron
Determine the number of moles of oxygen atoms in each of the following.
1) 4.93 mol H2O2
2) 2.01 mol N2O
Answer :
Part 1: 4.93 moles of [tex]H_2O_2[/tex] contains 9.86 moles of oxygen atoms.
Part 2: 2.01 moles of [tex]N_2O[/tex] contains 2.01 moles of oxygen atoms.
Explanation :
Part 1: 4.93 mol [tex]H_2O_2[/tex]
In 1 mole of [tex]H_2O_2[/tex], there are 2 atoms of hydrogen and 2 atoms of oxygen.
As, 1 mole of [tex]H_2O_2[/tex] contains 2 moles of oxygen atoms.
So, 4.93 moles of [tex]H_2O_2[/tex] contains [tex]4.93\times 2=9.86[/tex] moles of oxygen atoms.
Thus, 4.93 moles of [tex]H_2O_2[/tex] contains 9.86 moles of oxygen atoms.
Part 2: 2.01 mol [tex]N_2O[/tex]
In 1 mole of [tex]N_2O[/tex], there are 2 atoms of nitrogen and 1 atom of oxygen.
As, 1 mole of [tex]N_2O[/tex] contains 1 mole of oxygen atoms.
So, 2.01 moles of [tex]N_2O[/tex] contains [tex]2.01\times 1=2.01[/tex] moles of oxygen atoms.
Thus, 2.01 moles of [tex]N_2O[/tex] contains 2.01 moles of oxygen atoms.
Which is denser a liquid or solid why?
Answer:
Liquids are usually less dense than solids but more dense than air. Temperature can change a liquid's density. For example, increasing the temperature of water causes the molecules to spread farther apart. The farther apart the molecules are, the less dense the water is.
Answer:
Solids are usually much more dense than liquids and gases, but not always.
Explanation:
Mercury, a metallic element that is a liquid at room temperature, is denser than many solids. Aerogel, a very unusual human-made solid, is about 500 times less dense than wate
Forty milliliter of a liquid has a mass of 80 grams. What is the density of this substance?
Answer:
2.00g/ml
Explanation:
If water has a density of 1.00g/ml, and theres
40ml of it, it would weigh 40g
The substance is twice as dense as water, making its density 2.00g/ml
. A nail is hammered into a piece of wood. *
Answer:
and now you have a nail in your wood :)
Which of the following substances would have the greatest ductility?
A. Fe(s)
B. SiO2(s)
C. C(s)
D. NaCl(s)
Fe(s) would have the greatest ductility.
What is ductility?Ductility is the capability of a fabric to be drawn or plastically deformed without fracture. it's far therefore a demonstration of how 'gentle' or malleable the fabric is. The ductility of steel varies relying on the sorts and levels of alloying factors gift.
What are malleability and ductility?Ductility is the property of metallic associated with the capability to be stretched into twine without breaking. Malleability is the assets of metallic associated with the ability to be hammered into thin sheets without breaking. The outside force or strain is tensile pressure.
Learn more about ductility here https://brainly.com/question/496496
#SPJ2
CHEMISTRY!! 50 POINTS!
There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?
Answer:
The answer to this question is 33.8
What is an advantage of making plant-based products, such as cotton, instead of making petroleum-based products, such as plastics?
A Producing plant-based products requires more energy.
B Factories that make plant-based products generate more waste.
C Plants that are used to make the product can be replaced.
D All plant-based products cost more than petroleum-based products.
Answer:
Hey! sorry if im late but the
answer is :
C : Plants that are used to make the product can be replaced.
I hope this helps everyone out there!
Explanation:
i guessed lol
Answer:
C) Plants that are used to make the product can be replaced
Explanation:
Sorry I’m late I just did the test and got 100%
2-Methycyclohexanol is prepared commercially by catalytic hydrogenation of ocresol (2-methylphenol) and consists of a mixture of cis and trans isomers. (Note the spectrum is for the mixture.) (Hint: What do the peaks at 3.05 ppm and 3.75 ppm represent and what does their integration show
Answer:
Explanation:
From the information given :
The cis and trans isomer are in the ratio of 1:3, and the description of how the ratio of the cis- and trans-2-methylcyclohexanols from the 1H NMR spectrum can be explained as follows:
We are being told that the trans isomer's peak is 3.75 ppm (part per million). However, because the methyl group is far away from the OH group, it is less shilled than the cis isomer, which peaked at 3.05. Thus, the peak at 3.05 occurs in an area whose integration is three (3) times more than the peak at 3.75.
You need to make an aqueous solution of 0.192 M barium sulfide for an experiment in lab, using a 500 mL volumetric flask. How much solid barium sulfide should you add?
Answer:
Explanation:
Molecular weight of barium sulphide = 169
500 mL of .192 M barium sulfide = .5 x .192 moles of barium sulphide
= .096 moles of barium sulfide
= .096 x 169 gram of barium sulfide
= 16.22 grams of barium sulfide .
We shall have to add 16.22 gram .
Scientists are experimenting with pure samples of isotope X which is radioactive. The sample has a mass of 20. Grams. The half-life was measured to be 232 seconds. There is a second sample that weighs 80 grams. What is the half-life of the second sample
Answer:
Explanation:
Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .
Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .
A 50.0 g sample of an unknown substance, initially at 20.2 °C, was heated with 1.55 kJ of energy. The final temperature of the substance was 125.0 °C. Determine the specific heat of this substance.
Answer:
0.296j/g⁰c
Explanation:
we have the following information from this question before us.
mass iv substance = 50grams
we have initial temperature ti = 20.2⁰c
final temperature = 125⁰c
the energy that was provided = 155kj
we proceed with this formula
energy = mcΔT
1.55x10³ = 50 x c x (125-20.2)
1.55x10³ = c x 50gm x 104.8k
we divide through to get c
c = 1.55x10³/50g x 104.8
c = 0.296J/g⁰c
that is the specific heat of this substance.
thank you!