Answer:
She should not have multiplied the nitrogen atom by subscript 2.
Explanation:
Chemical formula:
3(NH₄)₂SO₄
Elements present in given formula:
Nitrogen
Hydrogen
Sulfur
Oxygen
Total number of atoms of elements:
N = 3×1×2 = 6
H = 4×2×3 = 24
S = 1×3 = 3
O = 3×4 = 12
The number nitrogen atoms are six. Elena did mistake by counting the number of nitrogen. She should didn't multiplied the nitrogen atom by subscript 2.
The first on is a and the second c I belive
Explanation:
To solve this problem, we need to understand the concept of precision and accuracy.
Through this understanding we apply to solve this problem.
- Precision is the ability to reproduce the same set of values in an experiment.
- Accuracy is the nearness or closeness of the measured value to the true value.
For example; if the true value is 28.0mm and the following readings were made:
A = 24.0mm, 24.0mm, 24.3mm, 23.9mm This is a precise measurement but not accurate.
B = 27.9mm, 28.0mm, 28.1mm: This is an accurate measurement because it is close to the true value.
Answer: 32.94 g
Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.
for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm
V = 15.0 L
T = 350.8 + 273 = 623.8 K
For krypton, 
n = 0.146 moles
for chlorine, 
n = 0.439
From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.
Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.
So, the amount of product formed is calculated from moles of krypton.
Molar mass of krypton tetrachloride is 225.61 gram per mol.
There is 1:1 mol ratio between krypton and krypton tetrachloride.

= 32.94 g of 
So, 32.94 g of the product will form.