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What property of gas molecule affects the speed at which it diffuses?

Tresset [83]

Answer:

Helium and Ethlyene Oxide. Your answer is MASS.

Explanation:

Diffusion is faster at higher temperatures because the gas molecules have greater kinetic energy. effusion refers to the movement of gas particles through a small hole. Graham's law state's that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles.

6 0

3 years ago

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Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the f

julia-pushkina [17]

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) $HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-$

$HNO_3$ is Bronsted lowry acid and $NO_3^-$ is its conjugate base.

$H_2O$ is Bronsted lowry base and $H_3O^+$ is its conjugate acid.

(b)

$CN^- + H_2O \rightarrow HCN + OH^-$

$CN^-$ is Bronsted lowry base and HCN is its conjugate acid.

$H_2O$ is Bronsted lowry acid and $OH^-$ is its conjugate base.

(c)

$H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-$

$H_2SO_4$ is Bronsted lowry acid and $HSO_4^-$ is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

$HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O$

$HSO_4^-$ is Bronsted lowry acid and $SO_4^{2-}$ is its conjugate base.

OH^- is Bronsted lowry base and $H_2O$ is its conjugate acid.

(e)

$O_{2-}+H_2O \rightarrow 2OH^-$

$O_{2-}$ is Bronsted lowry base and OH- is its conjugate acid.

$H_2O$ is Bronsted lowry acid and OH- is its conjugate base.

6 0

4 years ago

What volume of 3.00 MM HClHCl in liters is needed to react completely (with nothing left over) with 0.750 LL of 0.500 MM Na2CO3N

AlladinOne [14]

Answer: The volume of HCl needed is 0.250 L

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For sodium carbonate:

Molarity of sodium carbonate solution = 0.500 M

Volume of solution = 0.750 L

Putting values in above equation, we get:

$0.500M=\frac{\text{Moles of sodium carbonate}}{0.750}\\\\\text{Moles of sodium carbonate}=(0.500mol/L\times 0.750L)=0.375mol$

The chemical equation for the reaction of sodium carbonate and HCl follows:

$Na_2CO_3+2HCl\rightarrow 2NaCl+H_2CO_3$

By Stoichiometry of the reaction:

1 mole of sodium carbonate reacts with 2 moles of HCl

So, 0.375 moles of sodium carbonate will react with = $\frac{2}{1}\times 0.375=0.750mol$ of HCl

Now, calculating the volume of HCl by using equation 1:

Moles of HCl = 0.750 moles

Molarity of HCl = 3.00 M

Putting values in equation 1, we get:

$3.00M=\frac{0.750mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.750mol}{3.00mol/L}=0.250L$

Hence, the volume of HCl needed is 0.250 L

8 0

3 years ago

Identify two kinds of data the student will likely collect during his experiment. Identify the type of data collected and how th

-BARSIC- [3]

Qualitative (physical e. g. colour) and quantitative (numerical e. g. 3 grams) data

6 0

3 years ago

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A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr

umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0

4 years ago