<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
We are given:
Molarity of solution = 0.16 M
Given mass of manganese (II) nitrate tetrahydrate = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Putting values in above equation, we get:
Volume of water = Volume of solution = 398.4 mL
Hence, the amount of water required to prepare given amount of salt is 398.4 mL
0.091 moles are contained in 2.0 L of N2 at standard temperature and pressure.
Explanation:
Data given:
volume of the nitrogen gas = 2 litres
Standard temperature = 273 K
Standard pressure = 1 atm
number of moles =?
R (gas constant) = 0.08201 L atm/mole K
Assuming nitrogen to be an ideal gas at STP, we will use Ideal Gas law
PV = nRT
rearranging the equation to calculate number of moles:
PV = nRT
n = 
putting the values in the equation:
n = 
n = 0.091 moles
0.091 moles of nitrogen gas is contained in a container at STP.
Hydrogen and Oxygen by themselves have 1 and 6 valence electrons, respectively. 8 valence electrons is stable, so the atoms form bonds with each other to achieve 8 valence e-.
1 H atom + 1 H atom + 1 O atom = 8 valence e-
Iron oxide is rust. So oil would be an inhibitor.
Answer:
[Br₂] = 1.25M
Explanation:
2NO (g) + Br₂ (g) ⇄ 2NOBr (g)
Eq 0.80M ? 0.80M
That's the situation told, in the statement.
Let's make the expression for Kc
Kc = [NOBr]² / [Br₂] . [NO]²
Kc = 0.80² / [Br₂] . [0.80]²
0.80 = 1 / [Br₂]
[Br₂] = 1 / 0.80 → 1.25
