Answer:
The hydrogen produces the smaller amount of ammonia.
Step-by-step explanation:
We are given the masses of two reactants, so this is a <em>limiting reactant problem</em>.
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 28.02 2.016 17.03
N₂ + 3H₂ ⟶ 2NH₃
Mass/g: 70.0 7.00
1. Calculate the moles of N₂ and H₂
Moles N₂ = 70.0 × 1/28.02
Moles N₂ = 2.498 mol N₂
Moles H₂ = 7.00 × 2.016
Moles H₂ = 3.472 mol N₂
=====
2. Calculate the moles of NH₃ from each reactant
<em>From</em> N₂:
The molar ratio is 2 mol NH₃/1 mol N₂
Moles of NH₃ = 2.498 × 2/1
Moles of NH₃ = 4.996 mol NH₃
<em>From</em> H₂:
The molar ratio is 2 mol NH₃/3 mol H₂
Moles of NH₃ = 3.472 × 2/3
Moles of NH₃ = 4.139 mol NH₃
======
3. Identify the limiting reactant
The limiting reactant is H₂, because it produces fewer moles of NH₃.
Answer:
159 g OF LiNO2 WILL BE USED TO MAKE 0.25 L OF 0.75 M SOLUTION.
Explanation:
How many grams of LiNO2 are required to make 250 mL of 0.75 M?
First calculate the molarity in mol per dm3
0.75 M of LiNO2 reacts in 250 mL = 250 /1000 L volume
0.75 M = 0.25 L
In 1 L, the molarity of LiNO2 will be:
= (0.75 * 1/ 0.25) M
= 3 mol/dm3 of LiNO2
Next is to calculate the molarity in g/dm3:
Molarity in mol/dm3 = molarity in g/dm3 / RMM.
RMM of LiNO2
(Li = 7 , N =14 , 0 = 16)
RMM = ( 7 + 14 + 16 * 2) = 53 g/mol
Molarity in mol/dm3 = Molarity in g/dm3 / RMM
Molarity in g/dm3 = Molarity in mol/dm3 * RMM
Molarity in g/dm3 = 3 * 53
Molarity in g/dm3 = 159 g/dm3.
So therefore, to make 250 mL of 0.75 M of LiNO2, we use 159 grams of LiO2.
Answer:
A
Explanation:
The pesticides and chemicals in are food could get us seriously sick and cause food poisoning. We are hurting our water by dumping sewage water into lakes and rivers did you ever think that someday a river or a lake will mix in with the ocean.
Answer: 
Explanation:
Given
Initial mass 
half-life is 
At any time the left amount is given by
