How many moles are in 75.0g of Na2CO3

True or false. The greater the distance that the plane moves from an object, the lower the force that will be applied when the

Nadya [2.5K]

Answer:

I think its false-

Explanation:

4 0

3 years ago

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When lewis structures are drawn, do you only show valence electrons??

Lynna [10]

Yes you do cause its more explanation to

8 0

3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

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

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

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

How do you make a 3M solution of KBr? How many moles of KBr are 1 liter of a 0.3M solution?

pashok25 [27]

Molarity= moles/liter, so you would need 3mol KBr/1 liter
(0.3M)(1L)= 0.3mol KBr

8 0

3 years ago

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Draw one product that you would expect from the reaction of 1 mol of 1,3-butadiene and 1 mol of h2o, cat. h2so4?

Daniel [21]

1,3-butadiene is the simplest conjugated diene and undergoes 1,4 addition reaction in acidic environment.
Chemical reaction: CH₂=CH-CH=CH₂ + H₂O → CH₃-CH=CH-CH₂-OH.
CH₂=CH-CH=CH₂ - 1,3-butadiene.
CH₃-CH=CH-CH₂-OH - 2-buten-1-ol.
Diene or diolefin is a hydrocarbon that has two carbon double bonds.

3 0

3 years ago