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attashe74 [19]
3 years ago
5

Which mass of urea, CO(NH2)2, contains the same mass of nitrogen as 101.1g of potassium nitrate?

Chemistry
1 answer:
Effectus [21]3 years ago
4 0
In order to calculate the mass of nitrogen, we must first calculate the mass percentage of nitrogen in potassium nitrate. This is:
% nitrogen = mass of nitrogen / mass of potassium nitrate
% nitrogen = 14 / 101.1 x 100

The mass of nitrogen = % nitrogen x sample mass
= (14 / 101.1) x 101.1
= 14 grams

The molar weight of nitrogen is 14. Each mole of urea contains two moles of nitrogen. Therefore, for there to be 14 grams of nitrogen, there must be 0.5 moles of urea.
Mass of urea = moles urea x molecular weight urea
Mass of urea = 0.5 x 66.06
Mass of urea = 33.03 grams
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OLEGan [10]

Answer:

Reducing molecules.

Explanation:

NAD (Nicotinamide adenine dinucleotide) is the important molecule used by the living organisms for the generation of ATP. NADH is used almost in every biochemical cycle like glycolysis, kreb cycle and elelctron transport chain.

The NADH molecule is used as the reducing molecule in the biosynthesis of the different reaction. The NADH molecule reduces its hydrogen ions and also carry electrons for the synthesis of molecules. The NADH molecule is also used in the shuttle system as well.

Thus, the answer is reducing molecules.

6 0
3 years ago
State the number of protons contained in the atomic number 8<br>​
Nat2105 [25]

Answer:

There are 8 protons

Explanation:

{ \bf{atomic \: number = proton \: number}}

6 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]

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

<u>Explanation:</u>

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}}

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

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

<u>For hydrogen gas:</u>

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

  • <u>For nitrogen gas:</u>

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\%

  • <u>For hydrogen gas:</u>

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
Find the molecular mass of.
Margaret [11]

Answer:

  1. sodium oxide Nao
  2. aluminum chloride Al2cl3
  3. sodium sulphate Naso4
  4. <em>magnes</em><em>ium</em><em> </em><em>hydroxide</em><em> </em><em>Mgoh2</em>
3 0
3 years ago
Read 2 more answers
What is the freezing point in °C) of a 0.743 m
lesantik [10]

Answer:

-0.276 Degrees C

Explanation:

kf of water is 1.86

Freezing point depression= m x kf x i

i= ions present ( K+ Cl-) 1=2

1.86x2X.743= .276

Since its freezinf point depression the freezing point will lower

so

0-.276= -.276

7 0
3 years ago
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