Answer:
The number of moles of the gas is 9.295 moles or 9.30 moles
Explanation:
We use PV = nRT
Where P = 4.87 atm;
V = 67.54 L
R= 0.0821Latm/molK
T = 158 C = 158 +273 K = 431 K
the number of moles can be obtained by substituting the values in the respective columns and solve for n
n = PV / RT
n = 4.87 * 67.54 / 0.0821 * 431
n = 328.9198 / 35.3851
n = 9.295moles
The number of moles is approximately 9.30moles.
Basically all of the elements found in Group I of the periodic table also have this property. The ability to easily give up a single valence electron.
Answer:
<h2>2.49 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula
From the question we have
We have the final answer as
<h3>2.49 g/cm³</h3>
Hope this helps you
The flame goes an Orange-Red colour.
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
