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Shalnov [3]
3 years ago
5

What is the empirical formula for a compound that is 82.2% nitrogen and 17.8% hydrogen?

Chemistry
1 answer:
Stolb23 [73]3 years ago
3 0
Empirical formula=CH3
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To be honest, I learned this in school so I'll tell you XD

The formula of sodium oxide is Na2O
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Using the picture below, once air mass A reaches location B, the weather conditions at location B will most likely become
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I think the answer is a
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The volume of a sample of nitrogen is 6.00 liters at 35oC and 0.75 atm. What volume will it occupy at STP?
joja [24]
The  volume that  will  occupy  at STP  is  calculated   as  follows
by use  of ideal  gas  equation
that is  PV=nRT  where n  is  number of  moles  calculate  number of moles

n= PV/RT
p=0.75 atm
V=6.0  L
R = 0.0821  L.atm/k.mol
T=  35  +273= 308k
n=?

n=  (o.75  atm  x  6.0 L)/( 0.0821 L.atm/k.mol  x 308 k)=  0.178 moles

Agt  STP  1 mole=  22.4 L  what obout  0.178 moles

=  22.4  x0.178moles/ 1moles =3.98 L( answer C)
3 0
3 years ago
12. Why does a water molecule have a bent shape?
Andrej [43]

Answer:

A water molecule consists of two hydrogen atoms bonded to an oxygen atom, and its overall structure is bent. This is because the oxygen atom, in addition to forming bonds with the hydrogen atoms, also carries two pairs of unshared electrons. All of the electron pairs—shared and unshared—repel each other.

Explanation:

4 0
2 years ago
If a gas is 16 times as heavy as hydrogen which will diffuse faster and by what ratio​
vodka [1.7K]

Answer:

1. Hydrogen will diffuse faster.

2. The ratio of diffusion of hydrogen gas to that of the unknown gas is 4 : 1

Explanation:

Let the rate of diffusion of hydrogen gas, H2 be R1

Let the molar mass of H2 be M1

Let the rate of diffusion of the unknown gas be R2.

Let the molar mass of the unknown gas be M2.

Molar mass of H2 (M1) = 2x1 =2g/mol

Molar mass of unknown gas (M2) = 16 times that of H2

= 16 x 2 = 32g/mol

1. Determination of the gas that will diffuse faster. This is illustrated below:

R1/R2 = √(M2/M1)

R1/R2 = √(32/2)

R1/R2 = √16

R1/R2 = 4

Cross multiply

R1 = 4R2

From the above calculations, we can see that the rate of diffusion H2 (R1) is four times the rate of diffusion of the unknown gas (R2).

Therefore, hydrogen will diffuse faster.

2. Again, from the calculations made above, the ratio of diffusion of hydrogen (R1) to that of the unknown gas (R2) is given by;

R1/R2 = 4

Therefore, the ratio of diffusion of hydrogen (R1) to that of the unknown gas (R2) is:

4 : 1

7 0
3 years ago
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