The number of moles of hydrogen required will be 20 moles.
<h3>Stoichiometric mole ratio</h3>
First, let us look at the balanced equation of the reaction:
From the above equation, it is obvious that 1 mole of carbon requires 2 moles of hydrogen in order to produce a mole of methane.
In other words, the mole ratio of hydrogen to methane is 2:1. For every 1 mole of methane produced, 2 moles of hydrogen are consumed.
Now, what we want to produce is 10 moles of methane. The amount, in moles of hydrogen required, is calculated by:
10 moles x 2 = 20 moles.
Thus, 20 moles of hydrogen would be required to produce 10 moles of methane.
More on stoichiometric mole ratios can be found here: brainly.com/question/15053457
#SPJ1
Answer:
Heat flows from the reactor to the water
Explanation:
The thermal energy mentioned in the description is another way to say heat. The energy that is produced by the nuclear reactions leaves the reactor and enters the water, warming it.
The passage does <em>not </em>say that heat flows in the form of electricity, but rather that the turbines turned by the steam produce electricity.
The passage does <em>not </em>say that the steam produces the heat, but rather that the boiling of the water (caused by the heat) produces steam.
There are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.
<h3>HOW TO CALCULATE NUMBER OF MOLES?</h3>
The number of moles of a substance can be calculated by multiplying the molarity by the volume.
No. of moles = Molarity × volume
According to this question, 3L of a KBr solution are contained in a 0.4M.
no. of moles = 3L × 0.4M = 1.2moles
Therefore, there are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.
Learn more about no. of moles at: brainly.com/question/14919968
Answer:
Molarity of the solution = 3.000 M
Volume of the solution = 250.0 mL = 0.25 L
moles in 250.0 mL = molarity x volume of the solution
= 3.000 M x 0.25 L
= 0.75 mol
Hence, 0.75 mol of NaCl is needed to prepare 250.0 mL of 3.000 M NaCl solution.
Moles (mol) = mass (g) / molar mass (g/mol)
Moles of NaCl in 250.0 mL = 0.75 mol
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl in 250.0 mL = Moles x Molar mass
= 0.75 mol x 58.44 g/mol
= 43.83 g
Hence, 43.83 g of NaCl is needed to prepare 250.0 mL of 3.000 M solution.
Explanation:
