Answer:
A. Increasing the temperature will favor forward reaction and more CaCo3 formed.
B. More CaCo3 will be formed.
C. CaCo3 will decrease and more react ants formed.
D. Less CaCo3 will be formed.
E. Iridium is a catalyst so there is no effect
Explanation:
A. Temperature will increase because it's an endothermic reaction.
B. Adding Cao will favor forward reaction and more CaCo3 formed.
C. Removing methane, more react ants are formed and CaCo3 decreases.
D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.
The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.
There appears to be no table shown, but I can still answer. Aluminum has 3 valence electrons
The value of Kc for the equilibrium is 0.150 mole² / litre ²
<u>Explanation:</u>
<u>Given:</u>
An equilibrium mixture in an 1.00 L vessel contains 5.30 moles of
Mg(OH )₂ 0.800 moles of Mg²⁺ and 0.0010 moles OH₋
We have to find the value of Kc
- Step 1: Find the equilibrium Concentration.
- Step 2: Substitute the values in the equation.
- Step 3: Find the value of Kc.
- I have attached the document for the detailed explanation
The value of Kc for the equilibrium is 0.150 mole² / litre ²
