How will the volume of a gas be affected if the pressure is tripled, but the temperature remains the same?
2 answers:
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The answer for the following question is mentioned below.
+3
→2
+2
the equation is balanced.
Explanation:
The steps required to balance a chemical equation is:
Firstly,write the formula of the compound/substance.
Secondly,count the number of atoms in each element on each side of the equation
Thirdly,Adjust the co-efficient.Balance the atoms or the ions present in the greatest number.
You may do this by finding the lowest common multiple of two.
Continue systematically to balance the atoms or ions on the rest of the side (i.e.)on the other side.
Lastly,check the number of atoms each elements is same on both the sides
From the equation:
Before balancing the equation:
+
→
+
After balancing the equation:
+3
→2
+2
Therefore the given equation is balanced.
Answer:
Question 7 to 12 are given in attached file because character limit is only 5000
Explanation:
1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of sulfur. Calculate the empirical formula of this compound.
Given data:
Mass of sample = 15 g
Mass of sodium = 8.83 g
Mass of sulfur = 6.17 g
Empirical formula = ?
Solution:
Number of gram atoms of Na = 8.83 / 23 = 0.4
Number of gram atoms of S = 6.17 / 32 = 0.2
Atomic ratio:
Na : S
0.4/0.2 : 0.2/0.2
2 : 1
Na : S = 2 : 1
Empirical formula is Na₂S.
2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?
Given data:
Mass of phosphorus = 4.433 g
Mass of oxygen = 10.150 g - 4.433 g = 5.717 g
Empirical formula = ?
Solution:
Number of gram atoms of P = 4.433 / 30.9738 = 0.1431
Number of gram atoms of O = 5.717/ 15.999 = 0.3573
Atomic ratio:
P : O
0.1431/0.1431 : 0.3573/0.1431
1 : 2.5
P : O = 2(1 : 2.5)
Empirical formula is P₂O₅.
3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.
Given data:
Percentage of sodium = 36.48%
Percentage of sulfur = 25.41%
Percentage of oxygen = 38.11%
Empirical formula = ?
Solution:
Number of gram atoms of Na = 36.48 / 23 = 1.6
Number of gram atoms of S = 25.41/ 32 = 0.8
Number of gram atoms of O = 38.11/ 16 = 2.4
Atomic ratio:
Na : S : O
1.6/0.8 : 0.8/0.8 : 2.4/0.8
2 : 1 : 3
Na: S : O = 2 : 1 : 3
Empirical formula is Na₂SO₃.
4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.
Given data:
Percentage of iron = 63.52%
Percentage of sulfur = 36.48%
Empirical formula = ?
Solution:
Number of gram atoms of Fe = 63.52 / 55.845 = 1.14
Number of gram atoms of S = 36.48 / 32 = 1.14
Atomic ratio:
Fe : S
1.14/1.14 : 1.14/1.14
1 : 1
Fe : S = 1 : 1
Empirical formula is FeS.
5. Qualitative analysis shows that a compound contains 32.38%sodium, 22.65% sulfur and 44.99 % oxygen. Find the empirical formula of this compound.
Given data:
Percentage of sodium = 32.38%
Percentage of sulfur = 22.65%
Percentage of oxygen = 44.99%
Empirical formula = ?
Solution:
Number of gram atoms of Na = 32.38 / 23 = 1.4
Number of gram atoms of S = 22.65/ 32 = 0.7
Number of gram atoms of O = 44.99/ 16 = 2.8
Atomic ratio:
Na : S : O
1.4/0.7 : 0.7/0.7 : 2.8/0.7
2 : 1 : 4
Na: S : O = 2 : 1 : 4
Empirical formula is Na₂SO₄.
6. Analysis of a 20.0 gram sample of a compound containing only calcium and bromine indicates that 4.00 grams of calcium are present. What is the empirical formula of the compound formed?
Given data:
Mass of sample = 20g
Mass of bromine = 20 g - 4 g = 16 g
Mass of calcium = 4 g
Empirical formula = ?
Solution:
Number of gram atoms of bromine = 16 / 80= 0.2
Number of gram atoms of calcium = 4/ 40= 0.1
Atomic ratio:
Ca : Br
0.1/0.1 : 0.2/0.1
1 : 2
Ca: Br = 1 : 2
Empirical formula is CaBr₂.
