Answer:
When you balance a chemical equation you change coefficients.You never change subscripts. A coefficient is a whole number multiplier. To balance a chemical equation you add the coefficients to make sure there are the same number of atoms on each side of the arrow.
Explanation:
Hope this helps. MARK BRAINLEST!!
It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.
If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:
360 g : 1 L = x g : 30 mL
Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 30 mL
Now, crossing the products:
x </span>· 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000
x = 10.8 g
So, from 30 mL mixture, 10.8 g of NaCl could be extracted.
Let's calculate the same for 10 mL water instead of 30 mL.
360 g : 1 L = x g : 10 mL
Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 10 mL
Now, crossing the products:
x </span>· 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000
<span>x = 3.6 g
</span>
<span>So, from 10 mL mixture, 3.6 g of NaCl could be extracted.
</span>
Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and <span>from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
</span>3.6/10.8 = 1/3
Therefore, i<span>t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water. </span>
So it is able to reach out farther maybe?...
Answer:
Empirical Formula = NH₄NO₃ (Ammonium Nitrate)
Solution:
Step 1: Calculate Moles of each Element;
Moles of N = %N ÷ At.Mass of N
Moles of N = 35.0 ÷ 14
Moles of N = 2.5 mol
Moles of O = %O ÷ At.Mass of O
Moles of O = 59.96 ÷ 16
Moles of O = 3.7475 mol
Moles of H = [100% - (%N + %O)] ÷ At.Mass of H
Moles of H = [100% - (35.0 + 59.96)] ÷ 1.008
Moles of H = [100% - 94.96] ÷ 1.008
Moles of H = 5.04 ÷ 1.008
Moles of H = 5 mol
Step 2: Find out mole ratio and simplify it;
N H O
2.5 5 3.7475
2.5/2.5 5/2.5 3.7475/2.5
1 2 1.5
Multiply Mole Ratio by 2,
2 4 3
Result:
Empirical Formula = N₂H₄O₃
Or,
Empirical Formula = NH₄NO₃
This empirical formula is also a Molecular Formula for Ammonium Nitrate a well known Fertilizer and often misused in the formation of Explosives.
