Which of the following is NOT a strong electrolyte?
:

Answers:
Na2SO4
KI
CaCl2
LiOH
C6H1206

Answer:

Titanium [Ar] 3d2 4s2

Explanation:

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.

Answer:

you can see the empirical formula at the pic

The empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

Empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a compound.

(a) 1.245 g Ni : 5.381 g I

Mole of Ni ; Mole of I = 1.245/59 : 5.381/127 = 0.02 : 0.04 = 1:2

So the formula is NiI2

(b) 2.677 g Ba : 3.115 g Br

Mole of Ba : Mole of Br = 2.677/137 : 3.115/60 = 0.019 : 0.038

= 0.02 : 0.04 = 1:2

So the formula is BaBr2

(c) 2.128 g Be : 7.557 g S

Mole of Be : Mole of S = 2.128/9 : 7.557/32 = 0.2 : 0.2 = 1:1

So the formula is BeS

Thus, empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

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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!

Answer:

Each nerve cell is connected to the brain by electrical signals

Explanation:

Answer:

they said it above me

Explanation:

Answer:

chemotherapy

Explanation:

Answer:

solute

Explanation:

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

Answer:

true

hope this helps :)

Explanation:

True, [tex]BrF_5[/tex] (aq) is a good insulator.

A material or an object that does not easily allow heat, electricity, light, or sound to pass through it.

Bromine pentafluoride appears as a colourless, fuming liquid with a pungent odour. Used to make other chemicals and in rockets.

Hence,  [tex]BrF_5[/tex] (aq) is a good insulator.

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Answer:

Insoluble benzoic acid is formed

Explanation:

When benzoic acid is dissolved in sodium hydroxide solution, sodium benzoate is formed. This compound is highly soluble in the aqueous layer.

If this solution is now acidified, the equilibrum position shifts towards the undissociated benzoic acid and a precipitate appears.

An insoluble substance formed during crystallization or precipitation is called a precipitate. The precipitate of the reaction formed is of benzoic acid.

A precipitation reaction is a type of reaction in which an insoluble salt or substance is formed as the result of the reaction between the soluble salt solutions.

The substance or the salt produced that does not get dissolved in the solution is called a precipitate.

In the reaction, benzoic acid reacts with the sodium hydroxide salt and forms sodium benzoate salt as shown,

[tex]\rm C_{6}H_{5}COOH(s) + NaOH(aq) \rightarrow C_{6}H_{5}COO^{-}Na^{+}(aq) + H_{2}O(l)[/tex]

When the salt of sodium benzoate is acidified then the precipitation of benzoic acid is formed. The precipitate occurs because the equilibrium shifts towards the undissociated benzoic acid.

Therefore, a precipitate of benzoic acid is formed.

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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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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:

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.

This 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

[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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Answer:

0.42 g/cm3

Explanation:

Answer:

20/40=0.5 g/cm^3 becuase, mass/volume=density

Explanation:

Answer : The mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.

Explanation :

Balanced chemical reaction : It is a chemical reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

The balanced chemical reaction is:

[tex]PCl_3+Cl_2\rightarrow PCl_5[/tex]

By the stoichiometry of the reaction we can say that 1 mole of [tex]PCl_3[/tex] reacts with 1 mole of [tex]Cl_2[/tex] to give 1 mole of [tex]PCl_5[/tex].

From this we conclude that the mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.

Hence, the mole ratio of [tex]PCl_3[/tex] to [tex]PCl_5[/tex] is 1 : 1.

Answer:

A

Explanation:

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:

Therefore we can conclude that Option D does not represent an example of plants demonstrating the characteristics of life.

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Answer:

Waves is the disturbance or variation that transfer energy from one location to other

Answer:

this is not in my words but i think this will help

(put some of the words you would use in this)

Explanation:

Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves. ... Longitudinal waves are always characterized by particle motion being parallel to wave motion.

hope i helped

Answer:

true

Explanation:

Condensation is the opposite of vaporization. True. When does condensation occurs, does a gas lose or gain thermal energy?

Answer:

True

Explanation:

Condensation is formed with moisture and cold air, so the removal of thermal energy, or heat, would create the perfect conditions :)

Answer:The anwser is 5

Explanation:

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.

Answer:

C. molten rock below Earth’s surface

Explanation:

Answer:

[a]. 55.5kmol/min.

[b]. The average velocity = 8.3m/s.

pipe diameter = 1.0954m.

Explanation:

So, from the question above we are given the following data or information or parameters which are going to help us in solving the above question;

=> ''A gas stream that contains 1.50 mole% CO2 flows through a pipeline''

=> ''Twenty (20.0) kilograms of CO2 per minute is injected into the line.''

=> '' . A sample of the gas is drawn from a point in the line 150 meters downstream of the injection point and found to contain 2.3 mole% CO2''

=> ''molar gas density is 0.123k mol/m3''

[a].The gas flow rate (k mol/min) upstream of the injection point can be calculated as follows.

- Determine the mass balance= j₁ +[ 20 /.123 × 44] = j₃. Thus, the flow gas rate is given as;

w₁ ×  1.50 mole% CO2 + 3.6954= 2.3 mole% CO2 × w₃.

-3.610 = - 0.008w₁.

w₁ = 451.3 m³/min.

Thus, the flow rate = 0.123 × 451.3 = 55.51 kmol/min.

[b]. The average velocity = 150/ 18 seconds = 8.3 m/s.

[c]. Pipe Diameter can be calculated as follows;

0.123  × 8.3  × [3.12  × (pipe diameter)² / 4] = 55.51/60.

[1.0209  × 3.12]/4  × [pipe diameter]² = 0.9252.

0.8014065  ×  [pipe diameter]² = 0.9252.

[pipe diameter]² = 0.9252/ .8014065.

[pipe diameter]² = 1.15443.

[pipe diameter] = √1.15443.

[pipe diameter] = 1.2 m² = 1.0954m.

A) The gas flow rate (k.mol/min) upstream of the injection point is;

55.51 k.mol/min

B) The average velocity and the diameter of the pipe are respectively;

8.33 m/s and 1.072 m

Due to the fact that there is no reaction or accumulation in the pipeline, we can say that the input of CO₂ is equal to it's output. Thus;

Upstream + Injection = Downstream

A) If we assume that x is the flow rate of gas (kmol/min), then there is the we can write the balance as:

0.015x + (20/44) = 0.023(x + 20/44)

Expanding gives us;

0.015x + 0.4545 = 0.023x + 0.01045

Solving for x gives;

x = 55.51 kmol/min

B) Formula for velocity is;

v = distance/time

Thus, since the length of the pipe is 150 m/s and time is 18 seconds, then velocity is;

v = 150/18

v = 8.33 m/s

Now, volumetric flow rate is calculated from;

Q = q' * ρ

where;

q' is gas flow rate

ρ is gas density

Thus;

Q = 55.51 * 0.123

Q = 451.3 m³/min = 7.522 m³/s

We can also express average velocity as;

v = Q/A

where A is area = πD²/4

Thus;

8.33 = 7.522/(πD²/4)

Making D the subject of the formula gives;

D = √((4 * 7.522)/(8.33π))

D = 1.072 m

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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!

Answer:

um wheres thepassage?

Explanation:

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

Answer:

homogeneous is consisting a part of or people are similar to each other or are if same type.

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.

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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Answer:

What are we talking about here?

Explanation:

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].