Answer:
2NaOH + CO2 -> Na2CO3 + H2O
1) Find the moles of each substance
2) Determine the limitting reagent
∴ Carbon dioxide is limitting as it has a smaller value.
3) multiply the limiting reagent by the mole ratio of unknown over known
n(H2O ) = 0.3976369007 × 1/2
= 0.1988184504 moles
4) Multiply the number of moles by the molar mass of the substance.
m = 0.1988184504 × (1.008 × 2 + 16.00)
= 0.1988184504 × 18.016
= 3.581913202 g
Explanation:
Answer:
The answer is "
"
Explanation:
Please find the complete question in the attached file.
Equation:
at 
at equilibrium 
![p= 0.47 \ \ atm\\\\SO_2=3.3-0.47 = 2.83 \ \ atm\\\\O_2= 0.74 -\frac{0.47}{2}=0.74-0.235=0.555 \ atm\\\\K_P=\frac{[PSO_3]^2}{[PSO_2]^2[PO_2]}\\\\](https://tex.z-dn.net/?f=p%3D%200.47%20%5C%20%5C%20atm%5C%5C%5C%5CSO_2%3D3.3-0.47%20%3D%202.83%20%5C%20%5C%20atm%5C%5C%5C%5CO_2%3D%200.74%20-%5Cfrac%7B0.47%7D%7B2%7D%3D0.74-0.235%3D0.555%20%5C%20atm%5C%5C%5C%5CK_P%3D%5Cfrac%7B%5BPSO_3%5D%5E2%7D%7B%5BPSO_2%5D%5E2%5BPO_2%5D%7D%5C%5C%5C%5C)
Answer:
1.91 atm
Explanation:
Step 1: Calculate Henry's constant (k)
A gas has a solubility (C) of 2.45 g/L at a pressure (P) of 0.750 atm. These two variables are related to each other through Henry's law.
C = k × P
K = C/P
K = (2.45 g/L)/0.750 atm = 3.27 g/L.atm
Step 2: Calculate the pressure required to produce an aqueous solution containing 6.25 g/L of this gas at constant temperature.
We have C = 6.25 g/L and k = 3.27 g/L.atm. The required pressure is:
C = k × P
P = C/k
P = (6.25 g/L)/(3.27 g/L.atm) = 1.91 atm
Answer:
B. The s orbital and two p orbitals
Explanation:
