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3 years ago

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A gaseous mixture contains 2.00 moles of CO2 and 3.00 moles of O2. What percentage of the mixture’s mass is CO2?

1 answer:

Andrew [12]3 years ago

4 0

Answer:

Socratic app

Explanation:

it will help you

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a piece of candy is burned in a calorimeter raising up the tempature of 500g of water from 21 C to 25 C water has a specific hea

OverLord2011 [107]

The energy that was released by the candy is calculates using the below formula
Q=Mc delta T

Q= heat energy
m= mass (500g)
C= specific heat capacity) = 4.18 j/g/c
delta t =change in temperature = 25- 21 = 4 c

Q= 500 g x 4.18 j/g/c x 4c = 8360 j

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3 years ago

How do protons determine element identity

Butoxors [25]

Answer:

The number of protons also determines the identity of the element. ... Since the atom is electrically neutral, the number of electrons must equal the number of protons.

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3 years ago

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

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3 years ago

guys im morgz im going to put a nail on the seat in my mom room and im going to see how up her a- s s it will go lets see

vagabundo [1.1K]

Answer:

go outside and ponder your life choices dork.

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3 years ago

A gas containing 80% CH4 and 20% He is sent through a quart/ diffusion tube (see Figure P8.8) to recover the helium, Twenty perc

elena55 [62]

For a gas containing 80% CH4 and 20% He is sent through a quart diffusion tube, the composition is mathematically given as

%He=12.5%

%CH4=87.5%

<h3>What is the composition of the waste gas if 100 kg moles of gas are processed per minute?</h3>

Generally, the equation for the Material balance is mathematically given as

F=R+W

Therefore

100=0.20*1000+W

W=80kmol/min

In conclusion, waste gas compose

2.0/100*100=50/100*20+%*80

Hence

%He=12.5%

%CH4=87.5%

Read more about Chemical reaction

brainly.com/question/16416932

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2 years ago