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3 years ago

Can two different substances have the same percent composition by mass?

Chemistry

1 answer:

arlik [135]3 years ago

8 0

Answer:

Although the terms mass percent and percent composition are used interchangeably, they are different terms. The mass percent refers to the percentage of a component in a mixture, while the percent composition refers to the percentage of a particular chemical element in a mixture.

Explanation:

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What is the value of the van't Hoff factor for KCl if a 1.00m aqueous solution shows a vapor pressure depression of 0.734 mmHg a

yaroslaw [1]

<u>Answer:</u> The Van't Hoff factor for KCl is 1.74

<u>Explanation:</u>

We are given:

Molality of solution = 1 m

This means that 1 mole of a solute is present in 1 kg of solvent (water) or 1000 grams of water

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 1000 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=\frac{1000g}{18g/mol}=55.56mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of solute = 1 moles

Total moles = [1 + 55.56] = 56.56 moles

Putting values in above equation, we get:

$\chi_{(solute)}=\frac{1}{56.56}=0.0177$

The equation used to calculate relative lowering of vapor pressure follows:

$\frac{p^o-p_s}{p^o}=i\times \chi_{solute}$

where,

$\frac{p^o-p_s}{p^o}$ = relative lowering in vapor pressure = 0.734 mmHg

i = Van't Hoff factor = ?

$\chi_{solute}$ = mole fraction of solute = 0.0177

$p^o$ = vapor pressure of pure water = 23.76 torr

Putting values in above equation, we get:

$\frac{0.734}{23.76}=i\times 0.0177\\\\i=1.74$

Hence, the Van't Hoff factor for KCl is 1.74

7 0

3 years ago

The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: CH3CHF2(g) → C

Maru [420]

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $460^oC$ = $5.8\times 10^{-6}s^{-1}$

$K_2$ = rate constant at $T_2$ = $4\times K_1$

$Ea$ = activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $460^oC=273+460=733K$

$T_2$ = final temperature = ?

Now put all the given values in this formula, we get:

$\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]$

$T_2=791.1K$

Therefore, the final temperature would be, 791.1 K

4 0

3 years ago

Why is it that the shoreline tends to have milder temperatures in the winter than mainland?

Volgvan

I think its becuse the shoreline is shalow water that is warmed by the sun? ...... I may be rong

7 0

3 years ago

Handing out brainliest to the correct answer!!!

lora16 [44]

Answer -C They are defined by the number of electrons.
An orbital is a function that describes the probability of finding the electron with certain energy at certain distance from the nucleus therefore, it is defined by the number of electrons.

8 0

3 years ago

Read 2 more answers

I need chemistry help! How would I set up these problems?

Bond [772]

Answer: -

1) 8.33 minutes

2) 118.39 in/ s

180.43 m/min

10.83 km/ hr

Explanation: -

Speed of light = 3 x 10⁸ m/s