The answer to your question:
1.88 moles Na is 43.22 grams
If two different elements combine separately with a fixed mass of a third element, the ratio of the masses in which they do so are either the same as or a simple multiple of the ratio of the masses in which they combine with each other.
C. Enzyme because they catalyze biochemical reactions
First, we have to get:
1- The heat required to increase T of ice from -50 to 0 °C:
according to q formula:
q1 = m*C*ΔT
when m is the mass of ice = mol * molar mass
= 1 mol * 18 mol/g
= 18 g
and C is the specific heat capacity of ice = 2.09 J/g-K
and ΔT change in temperature = 0- (-50) = 50°C
by substitution:
∴q1 = 18 g * 2.09 J/g-K *50°C
= 1881 J = 1.881 KJ
2- the heat required to melt this mass of ice is :
q2 = n*ΔHfus
when n is the number of moles of ice = 1 mol
and ΔHfus = 6.01 KJ/mol
by substitution:
q2 = 1 mol * 6.01 KJ/mol
= 6.01 KJ
3- the heat required to increase the water temperature from 0°C to 60 °C is:
q3 = m*C*ΔT
when m is the mass of water = 18 g
C is the specific heat capacity of water = 4.18 J/g-K
ΔT is the change of Temperature of water = 60°C - 0°C = 60°C
by substitution:
∴q3 = 18 g * 4.18 J/g-K * 60°C
= 4514 J = 4.514 KJ
∴the total change of enthalpy = q1+q2+q3
= 1.881 KJ +6.01 KJ + 4.514 KJ
= 12.405 KJ
Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>
