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Gnesinka [82]
2 years ago
9

What is the molar mass of a gas if 0. 473 g of the gas occupies a volume of 376 ml at 23. 0°c and 1. 90 atm?.

Chemistry
1 answer:
frozen [14]2 years ago
6 0

Answer:

The molecular weight of the unknown gas is 16.1 g/mol.

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0.0278, how many significant figures
My name is Ann [436]

Answer:

There are three significant figures

Explanation:

When counting sig figs you don't count the zeros unless it is between a number greater than zero. The two zeros aren't between the greater numbers so there are only 3.

3 0
2 years ago
Why is it important to know where a resource, that you're using, is renewable or nonrenewable?
NemiM [27]

Answer:

It is important to know if a resource is renewable or non-renewable because renewable resources are self-sustaining (unlike non-renewable resources). If a non-renewable resource is used too often, that resource may eventually become non-existant. This can also happen to renewable resources, but it is more difficult.

3 0
2 years ago
The ksp of pbi2 is 1.4 x 10-8. what is the molar solubility of lead(ii iodide in a solution of 0.400 m sodium iodide?
tino4ka555 [31]
The solubility product of a substance us calculated by the product of the concentration of the dissociated ions in the solution raise to the stoichiometric coefficient of the ions. Therefore, we need the dissociation reaction. For this, it will have the reaction:

PbI2 = Pb^2+ + 2I-

We solve as follows:

Ksp = [Pb2+][I-]^2 = <span>1.4 x 10-8
</span><span>1.4 x 10-8 = x(2x)^2
</span><span>1.4 x 10-8 = 4x^3
x = 1.5x10^-3 M

The molar solubility would be </span>1.5x10^-3 M.
8 0
2 years ago
Ni-cad (nickel–cadmium) batteries have a slightly lower cell potential than the common alkaline value of 1.5 v normally used in
Lorico [155]
The half-reaction are:

Cd ---> Cd(OH)₂
The oxidation number of Cd changed from 0 to +2. So, the number of mol electron transferred here is 2.

NiO(OH) --> Ni(OH)₂
The oxidation number of Cd changed from +3 to +2. So, the number of mol electron transferred here is 1.

Now, the greatest common factor would be 2. So, we use n=2 for the formula for ΔG°. F is Faraday's constant equal to 96,485 J/mol e.

ΔG° = nFE° = (2)(96,485)(1.5) =<em> 289,455 J</em>


6 0
3 years ago
The half-life of phosphorus-32 is 14.30 days. how many milligrams of a 20.00 mg sample of phosphorus-32 will remain after 85.80
ZanzabumX [31]
The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.

             A(t) = A(o) (1/2)^(t/d)

where t is the certain period of time. Substituting the known values,

             A(t) = (20 mg)(1/2)^(85.80/14.30)

Solving,

           A(t) = 0.3125 mg

Hence, the answer is 0.3125 mg. 
7 0
3 years ago
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