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Zolol [24]
3 years ago
14

Liquid sample a has a higher vapor pressure than liquid sample b (both in closed containers). what can be said about sample a co

mpared with sample b?
Chemistry
1 answer:
Andreas93 [3]3 years ago
8 0
Vapor pressure is a criteria for a substance's volatility. This is the ability of a substance to transition from liquid to vapor. When the vapor pressure equals the external pressure, liquid turns to gas. Hence, if the substance has higher vapor pressure, then it is volatile. <em>Therefore, Sample A is more volatile than Sample B.</em>
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Calculate the atomic mass of oxygen if the three common isotopes of oxygen have masses of 15.995 amu (99.759% abundance), 16.995
Valentin [98]

Answer:

16 amu

Explanation:

6 0
3 years ago
What is the use of the crystallisation process?
ioda
  •  to obtain pure salt from seawater. 
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  • obtain pure alum crystals from an impure alum.

5 0
2 years ago
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Lastly, Snape thinks we should try one more calculation. What is the retention factor if the distance traveled by the solvent fr
Neporo4naja [7]

Answer:

0.1 is the retention factor.

Explanation:

Distance covered by solvent ,d_s= 2.0 cm

Distance covered by solute or ion,d = 0.20 cm

Retention factor(R_f) is defined as ratio of distance traveled by solute to the distance traveled by solvent.

R_f=\frac{d}{d_s}

R_f=\frac{0.20 cm}{2.0 cm}=0.1

0.1 is the retention factor.

7 0
3 years ago
Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. H+ + H2O2 ? H3
Margarita [4]

<u>Answer:</u> The rate law for the reaction is \text{Rate}=k'[H+][H_2O_2][Br^-]

<u>Explanation:</u>

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The chemical equation for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution follows:

2H^++2Br^-+H_2O_2\rightarrow Br_2+2H_2O

The intermediate reaction of the mechanism follows:

<u>Step 1:</u>  H^++H_2O_2\rightleftharpoons H_3O_2^+;\text{ (fast)}

<u>Step 2:</u>  H_3O_2^++Br^-\rightarrow HOBr+H_2O;\text{(slow)}

<u>Step 3:</u>  HOBr+H^++Br^-\rightarrow Br_2+H_2O;\text{(fast)}

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

\text{Rate}=k[H_3O_2^+][Br^-]          ......(1)

As, [H_3O_2^+] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for [H_3O_2^+] from step 1, we get:

K=\frac{[H_3O_2^+]}{[H^+][H_2O_2]}  

[H_3O_2^+]=K[H^+][H_2O_2]

Putting the value of [H_3O_2^+] in equation 1, we get:

\text{Rate}=k.K[H^+][H_2O_2][Br^-]\\\\\text{Rate}=k'[H+][H_2O_2][Br^-]

Hence, the rate law for the reaction is \text{Rate}=k'[H+][H_2O_2][Br^-]

4 0
3 years ago
Calculate the mass in grams for each of the following liquids.
WITCHER [35]
The density is calculated as mass per volume, so if we want to solve for mass, we would multiply density by volume.
For Part A: if we have a density of 0.69 g/mL, and a volume of 280 mL, multiplying these will give a mass of: (0.69 g/mL)(280 mL) = 193.2 g. Rounded to 2 significant figures, this is 190 g gasoline.
For Part B: if we have a density of 0.79 g/mL, and a volume of 190 mL, multiplying these will give a mass of: (0.79 g/mL)(190 mL) = 150.1 g. Rounded to 2 significant figures, this is equal to 150 g ethanol.
3 0
3 years ago
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