B will be the one I think tbh
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>
Answer:
Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.
Explanation:

Molarity of HCl solution = 0.1174 M
Volume of HCl solution = 83.15 mL = 0.08315 L
Moles of HCl = n



According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then 0.009762 mol of HCl will recat with:

Moles of Sodium carbonate = 0.004881 mol
Volume of the sodium carbonate containing solution taken = 1L
Concentration of sodium carbonate in the solution before the addition of HCl:
![[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D%5Cfrac%7B0.004881%20mol%7D%7B1%20L%7D%3D0.004881%20mol%2FL)
The solution would be like this for this specific problem:
<span>Given:
</span>66.0 g of carbon monoxide
reaction 2 C + O2 → 2 CO
<span>mol e= mass / molar mass <span>
<span>mole of 2CO = 66.0g / (12.011 15.999)g / mol </span>
mole of 2CO = 2.36 (CO and C has a 1:1 mole ratio)
mole of 2CO = 2.36 -> mole of 1 CO = 2.36 / 2 = 1.18
<span>We got 2 moles of C, thus 1.18 x 2 = 2.36
So, we 2.36 </span>moles of carbon are needed to produce 66.0 g of carbon monoxide in the </span>reaction
2 C + O2 → 2 CO.</span>
<span>To add, Carbon nonmetallic
and tetravalent, thus, making four electrons available to form covalent
chemical bonds. </span>