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3 years ago

9

Given two temperatures, 36 °F and 72 °F, which of the following correctly compares the number of opportunities for particles to

align during a chemical reaction?
Same at both
Greater at 36 °F
Greater at 72 °F
Does not depend on temperature

1 answer:

Liono4ka [1.6K]3 years ago

5 0

The correct answer is "Greater at 72 °F " hope it helps

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What can be said about a reaction with H = 62.4 kJ/mol and S = 0.145 kJ/(mol·K)?

Zigmanuir [339]

Answer:

At 430.34 K the reaction will be at equilibrium, at T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

Explanation:

1) Variables:

G = Gibbs energy
H = enthalpy
S = entropy

2) Formula (definition)

G = H + TS

=> ΔG = ΔH - TΔS

3) conditions

ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction

4) Assuming the data given correspond to ΔH and ΔS

ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K

=> T = [ΔH - ΔG] / ΔS

ΔG = 0 => T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K

This is, at 430.34 K the reaction will be at equilibrium, at T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

3 0

3 years ago

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18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0

3 years ago

What is the halogen in period 5?? it has six letters

Misha Larkins [42]

Simple,

take a look at your periodic table, if you have it labeled look at the Halogen

Group, it includes: Flourine, Chlorine, Bromine, Iodine, and Astatine.

Now, a period on the periodic table is read from left to right, and goes

down the rows of the periodic table.

Go to Period 5, go all the way to the Halogens, what is there?

Iodine.

Thus, your answer.

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3 years ago

The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution

Alex_Xolod [135]

The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is 2.64 x 10⁻⁵ M.

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3 years ago

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Vera_Pavlovna [14]

Answer: Should be A)

Explanation:

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3 years ago