Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, 
.
Explanation:
Given: Mass of methane = 146.6 g
As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.
The given reaction equation is as follows.
This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.
Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, .
Alkali are soluble bases, however not all bases are soluble in water, therefore not all bases are Alkali.
Answer:
2H+(aq) + 2OH-(aq) → 2H2O(l)
Explanation:
Step 1: The balanced equation
2HCl(aq)+Ca(OH)2(aq) → 2H2O(l)+CaCl2(aq)
This equation is balanced, we do not have the change any coefficients.
Step 2: The netionic equation
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.
2H+(aq) + 2Cl-(aq) + Ca^2+(aq) + 2OH-(aq) → 2H2O(l) + Ca^2+(aq) + 2Cl-(aq)
After canceling those spectator ions in both side, look like this:
2H+(aq) + 2OH-(aq) → 2H2O(l)
Answer:
45.02 L.
Explanation:
- Firstly, we need to calculate the no. of moles of water vapor.
- n = mass / molar mass = (36.21 g) / (18.0 g/mol) = 2.01 mol.
- We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.
<em><u>Using cross multiplication:</u></em>
1.0 mole of CO occupies → 22.4 L.
2.01 mole of CO occupies → ??? L.
∴ The volume of water vapor in 36.21 g = (22.4 L)(2.01 mole) / (1.0 mole) = 45.02 L.
Cs -137 has a half life of about 30 years. If 60 years pass, there is two half lives passed so 1/2 * 1/2= 1/4. Take 1/4 and multiply it with the mass given:
(1/4)*20mg=5mg left
