All categories

2 years ago

Based on the kinetic theory, which statement is true? (5 points)

Chemistry

1 answer:

Hoochie [10]2 years ago

7 0

Answer:

the particles of matter are arranged in different ways for the different states

Explanation:

because solid liquid and gas all three matters have different states for example the particles in a solid are closely packEd and form of movement is vibration

You might be interested in

Hello!

blondinia [14]

Answer:

$[CO]=[Cl_2]=0.01436M$

$[COCl_2]=0.00064M$

Explanation:

Hello there!

In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:

$K=\frac{[CO][Cl_2]}{[COCl_2]}$

Which can be written in terms of x, according to the ICE table:

$0.32=\frac{x^2}{0.015M-x}$

Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:

$[CO]=[Cl_2]=0.01436M$

$[COCl_2]=0.015M-0.01436M=0.00064M$

Regards!

5 0

3 years ago

2. Balance the equation below, and answer the following question: What volume of chlorine gas, measured at STP, is needed to com

Juliette [100K]

Balanced equation: 2Na(s) + Cl₂(g) ---> 2NaCl(s)

when we have STP conditions, we can use this conversion: 1 mol = 22.4 L

first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation.

molar mass of Na= 23.0 g/mol
ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)

calculations:

$6.25 g Na ( \frac{1 mol Na}{23.0 g} ) ( \frac{1 mol Cl_2}{2 mol Na} ) ( \frac{22.4 L}{1 mol} )= 3.04 L$

8 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago

The atomic mass of an element is found by which of the following methods?

Lera25 [3.4K]

On the periodic table it is the number on the bottom of the element.
If you know the amount of neutrons you can add it to the number of protons to find the atomic mass NUMBER, which is a good approximate of the atomic mass.

8 0

3 years ago

What is the molar mass of Agl2

Maru [420]

Answer:

107.8682

Explanation:

....................

7 0

3 years ago