Answer:
Number 1. The effective nuclear change of oxygen is greater that of fluorine.
A) 0.875 M of MgBr2 is 0.875 mol/L
B) 23.7 mL = 0.0237 L
0.875 mol => 1 L
x mol => 0.0237 L
Cross multiply
1x = 0.875 × 0.0237
x = 0.0207 mol
>> In 23.7 mL of 0.875 M MgBr2 solution there is 0.0207 moles of MgBr2
C) Molar mass of MgBr2 (*) = 24.305 + (2 × 79.904) = 184.113 g/mol
184.113 g => 1 mol
x g => 0.0207 mol
Cross multiply
1x = 184.113 × 0.0207
x = 3.8111 g
>> 0.0207 moles of MgBr2 is equivalent to 3.81 g of MgBr2
>> In 23.7 mL of 0.875 M MgBr2 solution there is 3.81 g of MgBr2
(*) Use your periodic table
Answer:
The equilibrium partial pressure of O2 is 0.545 atm
Explanation:
Step 1: Data given
Partial pressure of SO2 = 0.409 atm
Partial pressure of O2 = 0.601 atm
At equilibrium, the partial pressure of SO2 was 0.297 atm.
Step 2: The balanced equation
2SO2 + O2 ⇆ 2SO3
Step 3: The initial pressure
pSO2 = 0.409 atm
pO2 = 0.601 atm
pSO3 = 0 atm
Step 4: Calculate the pressure at the equilibrium
pSO2 = 0.409 - 2X atm
pO2 = 0.601 - X atm
pSO3 = 2X
pSO2 = 0.409 - 2X atm = 0.297
X = 0.056 atm
pO2 = 0.601 - 0.056 = 0.545 atm
pSO3 = 2*0.056 = 0.112 atm
Step 5: Calculate Kp
Kp = (pSO3)²/((pO2)*(pSO2)²)
Kp = (0.112²) / (0.545 * 0.297²)
Kp = 0.261
The equilibrium partial pressure of O2 is 0.545 atm
B. product
Think about it like this: a company produces products, not coefficients.