Chemical Name of Ag3PO4
1. Chemical Name of Ag3PO4
Answer:
Your mom
Explanation:
2. B. Solve the percentage composition of all atorns in compound1. AICI32. K2O3. NaBr4. Zn(OH)25. Ag3PO4
Solution (1)
Step 1: Calculate the molar mass of AlCl₃.
molar mass = (26.98 g/mol × 1) + (35.45 g/mol × 3)
molar mass = 133.33 g/mol
Step 2: Calculate the percentage composition of the elements in AlCl₃.
For %Al
[tex]\text{\%Al} = \frac{\text{26.98 g/mol × 1}}{\text{133.33 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%Al = 20.24\%}}[/tex]
For %Cl
[tex]\text{\%Cl} = \frac{\text{35.45 g/mol × 3}}{\text{133.33 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%Cl = 79.76\%}}[/tex]
Solution (2)
Step 1: Calculate the molar mass of K₂O.
molar mass = (39.10 g/mol × 2) + (16.00 g/mol × 1)
molar mass = 94.20 g/mol
Step 2: Calculate the percentage composition of the elements in K₂O.
For %K
[tex]\text{\%K} = \frac{\text{39.10 g/mol × 2}}{\text{94.20 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%K = 83.01\%}}[/tex]
For %O
[tex]\text{\%O} = \frac{\text{16.00 g/mol × 1}}{\text{94.20 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%O = 16.99\%}}[/tex]
Solution (3)
Step 1: Calculate the molar mass of NaBr.
molar mass = (22.99 g/mol × 1) + (79.90 g/mol × 1)
molar mass = 102.89 g/mol
Step 2: Calculate the percentage composition of the elements in NaBr.
For %Na
[tex]\text{\%Na} = \frac{\text{22.99 g/mol × 1}}{\text{102.89 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%Na = 22.34\%}}[/tex]
For %Br
[tex]\text{\%Br} = \frac{\text{79.90 g/mol × 1}}{\text{102.89 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%Br = 77.66\%}}[/tex]
Solution (4)
Step 1: Calculate the molar mass of Zn(OH)₂.
molar mass = (65.39 g/mol × 1) + (16.00 g/mol × 2) + (1.008 g/mol × 2)
molar mass = 99.406 g/mol
Step 2: Calculate the percentage composition of the elements in Zn(OH)₂.
For %Zn
[tex]\text{\%Zn} = \frac{\text{65.39 g/mol × 1}}{\text{99.406 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%Zn = 65.78\%}}[/tex]
For %O
[tex]\text{\%O} = \frac{\text{16.00 g/mol × 2}}{\text{99.406 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%O = 32.19\%}}[/tex]
For %H
[tex]\text{\%H} = \frac{\text{1.008 g/mol × 2}}{\text{99.406 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%O = 2.03\%}}[/tex]
Solution (5)
Step 1: Calculate the molar mass of Ag₃PO₄.
molar mass = (107.87 g/mol × 3) + (30.97 g/mol × 1) + (16.00 g/mol × 4)
molar mass = 418.58 g/mol
Step 2: Calculate the percentage composition of the elements in Ag₃PO₄.
For %Ag
[tex]\text{\%Ag} = \frac{\text{107.87 g/mol × 3}}{\text{418.58 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%Ag = 77.31\%}}[/tex]
For %P
[tex]\text{\%P} = \frac{\text{30.97 g/mol × 1}}{\text{418.58 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%P = 7.40\%}}[/tex]
For %O
[tex]\text{\%O} = \frac{\text{16.00 g/mol × 4}}{\text{418.58 g/mol}} × 100[/tex]
[tex]\boxed{\text{\%O = 15.29\%}}[/tex]
#LearnMore
3. B. Solve the percentage composition of all atoms in a compound1. AICI32. K203. NaBr4. Zn(OH)25. Ag3PO4
Solution (1)
Step 1: Calculate the molar mass of AlCl₃.
molar mass = (26.98 g/mol × 1) + (35.45 g/mol × 3)
molar mass = 133.33 g/mol
Step 2: Calculate the percentage composition of AlCl₃.
For mass percent of Al
[tex]\text{mass percent of Al} = \frac{\text{26.98 g/mol × 1}}{\text{133.33 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of Al = 20.24\%}}[/tex]
For mass percent of Cl
[tex]\text{mass percent of Cl} = \frac{\text{35.45 g/mol × 3}}{\text{133.33 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of Cl = 79.76\%}}[/tex]
Solution (2)
Step 1: Calculate the molar mass of K₂O.
molar mass = (39.10 g/mol × 2) + (16.00 g/mol × 1)
molar mass = 94.20 g/mol
Step 2: Calculate the percentage composition of K₂O.
For mass percent of K
[tex]\text{mass percent of K} = \frac{\text{39.10 g/mol × 2}}{\text{94.20 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of K = 83.01\%}}[/tex]
For mass percent of O
[tex]\text{mass percent of O} = \frac{\text{16.00 g/mol × 1}}{\text{94.20 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of O = 16.99\%}}[/tex]
Solution (3)
Step 1: Calculate the molar mass of NaBr.
molar mass = (22.99 g/mol × 1) + (79.90 g/mol × 1)
molar mass = 102.89 g/mol
Step 2: Calculate the percentage composition of NaBr.
For mass percent of Na
[tex]\text{mass percent of Na} = \frac{\text{22.99 g/mol × 1}}{\text{102.89 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of Na = 22.34\%}}[/tex]
For mass percent of Br
[tex]\text{mass percent of Br} = \frac{\text{79.90 g/mol × 1}}{\text{102.89 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of Br = 77.66\%}}[/tex]
Solution (4)
Step 1: Calculate the molar mass of Zn(OH)₂.
molar mass = (65.39 g/mol × 1) + (16.00 g/mol × 2) + (1.008 g/mol × 2)
molar mass = 99.41 g/mol
Step 2: Calculate the percentage composition of Zn(OH)₂.
For mass percent of Zn
[tex]\text{mass percent of Zn} = \frac{\text{65.39 g/mol × 1}}{\text{99.41 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of Zn = 65.78\%}}[/tex]
For mass percent of O
[tex]\text{mass percent of O} = \frac{\text{16.00 g/mol × 2}}{\text{99.41 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of O = 32.19\%}}[/tex]
For mass percent of H
[tex]\text{mass percent of H} = \frac{\text{1.008 g/mol × 2}}{\text{99.41 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of H = 2.03\%}}[/tex]
Solution (5)
Step 1: Calculate the molar mass of Ag₃PO₄.
molar mass = (107.9 g/mol × 3) + (30.97 g/mol × 1) + (16.00 g/mol × 4)
molar mass = 418.7 g/mol
Step 2: Calculate the percentage composition of Ag₃PO₄.
For mass percent of Ag
[tex]\text{mass percent of Ag} = \frac{\text{107.9 g/mol × 3}}{\text{418.7 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of Ag = 77.31\%}}[/tex]
For mass percent of P
[tex]\text{mass percent of P} = \frac{\text{30.97 g/mol × 1}}{\text{418.7 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of P = 7.40\%}}[/tex]
For mass percent of O
[tex]\text{mass percent of O} = \frac{\text{16.00 g/mol × 4}}{\text{418.7 g/mol}} × 100[/tex]
[tex]\boxed{\text{mass percent of O = 15.29\%}}[/tex]
#CarryOnLearning
4. Activity 5: Know my percentage!A. Solve the percentage composition of an underlined atom being asked in a compound.1. Na2S2. Fe(OH)23. CaSO44. Al(HCO3)35. Zn(ClO2)2B. Solve the percentage composition of all atoms in a compound1. AlCl32. K2O3. NaBr4. Zn(OH)25. Ag3PO4
Answer:
Hopefully, makatulong to sa pag-aaral niyo. ❤️
A. Solve the Underlined words
1. Na2S
% age of Na= Molar Mass of Na x Subscript of Na/Molar Mass of Na2S x 100
= 23 g of Na x 2 Numbers of Na/(23 g of Na x 2)+ (32g of S x 1) x 100
= 46 g Na/78 g Na2S x 100
= 0.589 x 100
= 58.9% of Na
2. Fe(OH)2
% age of O= Molar Mass of O x Subscript of O/Molar Mass of Fe(Oh)2 x 100
= 16 g of Fe x 2 Number/s of Fe/(56 g of Fe x 1)+(16 g of O x 2)+(1 g of H x 2) x 100
= 32 g O/90 g Fe(OH)2 x 100
= 0.355 x 100
= 35.5% of O
3. CaSO4
% age of Ca= Molar Mass of Ca x Subscript of Ca/Molar Mass of CaSO4 x 100
= 40 g of Ca x 1 Number of Ca/(40 g of Ca x 1)+(31 g of S x 1)+(16 g of O x 4) x 100
= 40 g Ca/136 g CaSO4 x 100
= 0.294 x 100
= 29.4 % of Ca
4. Al (HCO3)3
% age of H= Molar Mass of H x Subscript of H/Molar Mass of Al (HCO3)3 x 100
= 1 g of H x 3 Number/s of H/(27 g of Al x 1)+(1 g of H x 3)+(12 g of C x 3)+ (48g of O x 3) x 100
= 3 g Al/210 g Al (HCO3)3 x 100
= 0.014 x 100
= 1.4 % of H
5. Zn(ClO2)2
% age of O= Molar Mass of O x Subscript of O/Molar Mass of Zn(ClO2)2 x 100
= (16 g of O x 2) x 2 Number/s of O/(65 g of Zn x 1)+(35 g of Cl x 2)+(32 g of O x 2) x 100
= 32 g of O x 2 Numbers of O/(65 g of Zn x 1)+(35 g of Cl x 2)+(32 g of O x 2) x 100
= 64 g O/199 g Zn(ClO2)2 x 100
= 0.322 x 100
= 32.2 % of O
B.
1. AlCl3
% age of Al= Molar Mass of Al x Subscript of Al/Molar Mass of AlCl3 x 100
= 27 g of Al x 1 Number of Al/(27 g of Al x 1)+(35 g of Cl x 3) x 100
= 27 g Al/132 g AlCl3 x 100
= 0.204 x 100
= 20.4 % of Al
% age of Cl= Molar Mass of Cl x Subscript of Cl/Molar Mass of AlCl3 x 100
= 35 g of Cl x 3 Number/s of Cl/(27 g of Al x 1)+(35 g of Cl x 3) x 100
= 105 g Cl/132 g AlCl3 x 100
= 0.795 x 100
= 79.5 % of Cl
Total Percentage:
Al = 20.4%
Cl =+ 79.5%
= 99.9% or 100%
2. K2O
% age of K= Molar Mass of K x Subscript of K/Molar Mass of K2O x 100
= 39 g of K x 2 Number/s of K/(39 g of Al x 2)+(16 g of O x 1) x 100
= 78 g K/94 g K2O x 100
= 0.829 x 100
= 82.9 % of K
% age of O= Molar Mass of O x Subscript of O/Molar Mass of K2O x 100
= 16 g of K x 1 Number of O/(39 g of Al x 2)+(16 g of O x 1) x 100
= 16 g O/94 g K2O x 100
= 0.170 x 100
= 17.0 % of O
Total Percentage:
K = 82.9%
O =+ 17.0%
= 99.9% or 100%
