Yeah, it would be B (1.2 x 102 m^3) because the measurement gave it away even though other numbers were higher, however, the measurements for those were smaller in size.
Answer:
the volume occupied by 3.0 g of the gas is 16.8 L.
Explanation:
Given;
initial reacting mass of the helium gas, m₁ = 4.0 g
volume occupied by the helium gas, V = 22.4 L
pressure of the gas, P = 1 .0 atm
temperature of the gas, T = 0⁰C = 273 K
atomic mass of helium gas, M = 4.0 g/mol
initial number of moles of the gas is calculated as follows;

The number of moles of the gas when the reacting mass is 3.0 g;
m₂ = 3.0 g

The volume of the gas at 0.75 mol is determined using ideal gas law;
PV = nRT

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.
It dissolves I think I know I am expert and. Ute
<span>2.51 grams
You want to prepare 19.16 g of some solution which will have 13.1% of it's mass being sucrose. So we just need to perform some simple multiplication:
19.16g * 0.131 = 2.50996g
Rounding to 3 significant figures gives 2.51 g.</span>
Answer : The mass defect required to release energy is 6111.111 kg
Explanation :
To calculate the mass defect for given energy released, we use Einstein's equation:

E = Energy released = 
= mass change = ?
c = speed of light = 
Now put all the given values in above equation, we get:


Therefore, the mass defect required to release energy is 6111.111 kg