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Blababa [14]
3 years ago
15

Which of the following is a qualitative physical property?

Chemistry
1 answer:
saw5 [17]3 years ago
5 0

Answer:

c

Explanation:

the rest are chemical properties

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Which form of energy can be connected closely to climate change?
Art [367]

Answer:

COAL

Explanation:

4 0
3 years ago
2) A common "rule of thumb" -- for many reactions around room temperature is that the
babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$  and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}} $

$\Rightarrow \frac{K_1}{K_2}=  e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

           R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}=  3.479 \times 10^{-6}$

$\Rightarrow  \frac{K_2}{K_1}=  e^{3.479 \times 10^{-6}}$

$\Rightarrow  \frac{K_2}{K_1}=  1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0
2 years ago
Consider two gases, A and B, are in a container at room temperature. What effect will the following changes have on the rate of
KengaRu [80]

Answer:

i think it will increase the rate of chemical reaction as pressure is directly proportional to the reactivity of gas.

8 0
3 years ago
Read 2 more answers
A mixture of gases has 0.3000 mol of CO2, 0.2706 mol of SO2 and 0.3500 mol of water vapor. The total pressure of the mixture is
notka56 [123]

Answer:

Partial pressure SO₂ → 0.440 atm

Explanation:

We apply the mole fraction concept to solve this:

Moles of gas / Total moles = Partial pressure of the gas / Total pressure

Total moles = 0.3 moles of CO₂ + 0.2706 moles of SO₂ + 0.35 moles H₂O

Total moles = 0.9206 moles

Mole fraction SO₂ = 0.2706 moles / 0.9206 moles → 0.29

Now, we can know the partial pressure:

0.29 = Partial pressure SO₂ / Total pressure

0.29 = Partial pressure SO₂ / 1.5 atm

0.29 . 1.5atm = Partial pressure SO₂ → 0.440 atm

8 0
3 years ago
HOW TO DO STOICHIOMETRY​
stellarik [79]

Answer:

1.Balance the equation.

2.Convert units of a given substance to moles.

3.Using the mole ratio, calculate the moles of substance yielded by the reaction.

4.Convert moles of wanted substance to desired units.

Explanation:

8 0
2 years ago
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