Chemistry Topics
Answer:
1) CuBr2
Total mass = 223.35 g/mol
Cu = 28.45%
Br = 71.55%
2) NaOH
Total mass = 40 g/mol
Na = 57.48%
O = 40%
H = 2.52%
3) KMnO4
K = 24.74%
Mn = 34.76%
O4 = 40.50%
4) NH4CO3
N = 17.95%
H4 = 5.18%
C = 15.39%
O3 = 61.49%
5) NH4C1
N = 46.61%
H = 13.44%
C1 = 39.95%
Solution:
1) CuBr2
Cu = 63.55 g
Br = 79.90 x 2 = 159.8 g
159.8 + 63.55 = 223.35 g/ mol
[tex]Cu = \frac{63.55}{223.35} times 100= 28.45%[/tex]
[tex]Br = \frac{159.8}{223.35} times 100 = 71.55[/tex]
Check: 28.45 + 71.55 = 100%
2) NaOH
Na = 22.99 g
O = 16 g
H = 1.01 g
22.99 + 16 + 1.01 = 40 g/mol
[tex]Na = \frac{22.99}{40} times 100 =57.48[/tex]
[tex]O = \frac{16}{40} times 100=40[/tex]
[tex]H= \frac{1.01}{40} times 100=2.52[/tex]
Check: 40 + 57.48 + 2.52 = 100%
3) KMnO4
K = 39.10 g
Mn = 54.94 g
O = 64 g
39.10 + 54.94 + 64 = 158.04 g/mol
[tex]K =\frac{39.10}{158.04} times 100 = 24.74\\[/tex]
[tex]Mn = \frac{54.94}{158.04} times 100=34.76[/tex]
[tex]O = \frac{64}{158.04} times 100 = 40.50[/tex]
Check: 24.74 + 34.76 + 40.50 = 100
4) NH4CO3
N= 14.01 g
H4=4.04 g
C= 12.01 g
O3= 48 g
= 78.06 g/mol
[tex]N = \frac{14.01}{78.06} times 100=17.95\\H4= \frac{4.04}{78.06} times 100= 5.18\\C = \frac{12.01}{78.06} times 100 = 15.39\\O3 = \frac{48}{78.06} times 100 = 61.49[/tex]
5) NH4C1
N = 14.01 g
H4 = 4.04 g
C = 12.01 g
= 30.06 g/mol
[tex]N = \frac{14.01}{30.06} times 100 = 46.61\\H4 = \frac{4.04}{30.06}times 100 = 13.44\\ C = \frac{12.01}{30.06} times 100 = 39.95[/tex]
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