24. Which description correctly identifies the substance below?*

The graph shows the changes in the phase of ice when it is heated.

kow [346]

Answer:

A) 0 °C, because it is the melting point of ice.

Explanation:

Point B is the temperature at which the water is converted from ice (solid phase) to liquid water (liquid phase), which is the melting transition of water.

Melting point of the water is at 0.0°C.

<em>So, the right choice is: A) 0 °C, because it is the melting point of ice. </em>

6 0

3 years ago

Read 2 more answers

Pls it’s urgent

Rainbow [258]

Answer:

0.02 moles.

Explanation:

volume of H₂ gas at R.T.P = 480 cm³

Where

R.T.P = room temperature and pressure

molar volume of gas at = 24000 cm³

no. of moles of hydrogen = ?

Solution:

formula Used

no. of moles = volume of gas / molar volume

put values in above equation

no. of moles = 480 cm³ / 24000 cm³/mol

no. of moles = 0.02 mol

So,

no. of moles of hydrogen in 480 cm³ is 0.02 moles.

4 0

2 years ago

Describe how oxidation and reduction involve electrons, change oxidation numbers, and combine in

Sholpan [36]

Answer:

Redox

Explanation:

Reduction is gain of electrons

oxidation is loss of electrons

3 0

3 years ago

Never mind wrong question

Leokris [45]

Its okay my friend. you dont need to over stress it.

7 0

3 years ago

Read 2 more answers

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