Solve for the percentage composition of the following compounds., 1) CuBr2, 2) Na0H, 3) KMn04, 4) NH4CO3, 5) NH4C1, please i need it now!, warning: nonsense = report!, related/correct answer = brainliest, Please show your solutions,, Thank you!

24 sec read


Solve for the percentage composition of the following compounds.


1) CuBr2
2) Na0H
3) KMn04
4) NH4CO3
5) NH4C1

please i need it now!

warning: nonsense = report!
related/correct answer = brainliest

Please show your solutions,
Thank you!

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

Cu = \frac{63.55}{223.35} times 100= 28.45%

Br = \frac{159.8}{223.35} times 100 = 71.55

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

Na = \frac{22.99}{40} times 100 =57.48

O = \frac{16}{40} times 100=40

H= \frac{1.01}{40} times 100=2.52

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

K =\frac{39.10}{158.04} times 100 = 24.74\\

Mn = \frac{54.94}{158.04} times 100=34.76

O = \frac{64}{158.04} times 100 = 40.50

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

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

5) NH4C1

N = 14.01 g

H4 = 4.04 g

C = 12.01 g

= 30.06 g/mol

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

Chemistry made Easy!

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#CarryOnLearning

Cheers!

-InfinityER Original Copy


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