Answer:
Hello some parts of your question is missing
Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,000 and 100,000 g mol−1 as indicated below: (a) Equal numbers of molecules of each sample (b) Equal masses of each sample (c) By mixing in the mass ratio 0.145:0.855 the two samples with molar masses of 10,000 and 100,000 g mol−1 For each of the mixtures, calculate the number-average and weight-average molar masses and comment upon the meaning of the values.
Answers:
a) 46.666.66 g/mol
b) 20930.23 g/mol
c)43333.33 g/mol
Explanation:
A)The equal number of molecules of each sample can be calculated using 
because for the number of molecules to be equal : n1 = n2 = n3 = n
Mn = 46666.66 g/mol
B ) To calculate the equal masses of each sample
we apply this equation

ATTACHED BELOW IS THE REMAINING PART OF THE SOLUTION
Search up “Chemistry, how to do empirical formulas” a guy named Tyler Dewitt is really good at explaining
Answer:
Cu(NO3)2 --> MM187.5558
NiNO3 *COEF2* --> 120.6983
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit