26. 12h₂o h = 24 o = 12
27. 5n₂o₄ n = ____ o = ____
28. 3clf cl = ____ f = ____
29. 7p₂o₅ p = ____ o = ____
30. 2krcl₆ kr = ____ cl = ____
31. 5al(c₂h₃o₂)₃ al = ____ c = ____ h = ____ o = ____
32. 3(nh₄)₂cr₂o₇ n = ____ h = ____ cr = ____ o = ____
33. 5fe₃(po₄)₂ fe = ____ p = ____ o = ____
34. 2nh₄no₃ n = ____ h = ____ o = ____
35. 5bac₄h₄o₆ ba = ____ c = ____ h = ____ o = ____
36. 4cu(hso₃)₂ cu = ____ h = ____ s = ____ o = ____
37. 9au(no₂)₂ au = ____ n = ____ o = ____
38. 3k₂zno₂ k = ____ zn = ____ o = ____
39. 3sr(mno₄)₂ sr = ____ mn = ____ o = ____
40. 4al₂(co₃)₃ al = ____ c = ____ o = ____

Question

26. 12h₂o  h = 24  o = 12
27. 5n₂o₄  n = ____  o = ____
28. 3clf  cl = ____  f = ____
29. 7p₂o₅  p = ____  o = ____
30. 2krcl₆  kr = ____  cl = ____
31. 5al(c₂h₃o₂)₃  al = ____  c = ____  h = ____  o = ____
32. 3(nh₄)₂cr₂o₇  n = ____  h = ____  cr = ____  o = ____
33. 5fe₃(po₄)₂  fe = ____  p = ____  o = ____
34. 2nh₄no₃  n = ____  h = ____  o = ____
35. 5bac₄h₄o₆  ba = ____  c = ____  h = ____  o = ____
36. 4cu(hso₃)₂  cu = ____  h = ____  s = ____  o = ____
37. 9au(no₂)₂  au = ____  n = ____  o = ____
38. 3k₂zno₂  k = ____  zn = ____  o = ____
39. 3sr(mno₄)₂  sr = ____  mn = ____  o = ____
40. 4al₂(co₃)₃  al = ____  c = ____  o = ____

Accepted Answer

### Problem 26: $12\ce{H_2O}$
# Explanation:
## Step1: Calculate H atoms
In one $\ce{H_2O}$, there are 2 H atoms. For 12 molecules, $12\times2 = 24$.
## Step2: Calculate O atoms
In one $\ce{H_2O}$, there is 1 O atom. For 12 molecules, $12\times1 = 12$.
# Answer:
H = 24, O = 12

### Problem 27: $5\ce{N_2O_4}$
# Explanation:
## Step1: Calculate N atoms
In one $\ce{N_2O_4}$, there are 2 N atoms. For 5 molecules, $5\times2 = 10$.
## Step2: Calculate O atoms
In one $\ce{N_2O_4}$, there are 4 O atoms. For 5 molecules, $5\times4 = 20$.
# Answer:
N = 10, O = 20

### Problem 28: $3\ce{ClF}$
# Explanation:
## Step1: Calculate Cl atoms
In one $\ce{ClF}$, there is 1 Cl atom. For 3 molecules, $3\times1 = 3$.
## Step2: Calculate F atoms
In one $\ce{ClF}$, there is 1 F atom. For 3 molecules, $3\times1 = 3$.
# Answer:
Cl = 3, F = 3

### Problem 29: $7\ce{P_2O_5}$
# Explanation:
## Step1: Calculate P atoms
In one $\ce{P_2O_5}$, there are 2 P atoms. For 7 molecules, $7\times2 = 14$.
## Step2: Calculate O atoms
In one $\ce{P_2O_5}$, there are 5 O atoms. For 7 molecules, $7\times5 = 35$.
# Answer:
P = 14, O = 35

### Problem 30: $2\ce{KrCl_6}$
# Explanation:
## Step1: Calculate Kr atoms
In one $\ce{KrCl_6}$, there is 1 Kr atom. For 2 molecules, $2\times1 = 2$.
## Step2: Calculate Cl atoms
In one $\ce{KrCl_6}$, there are 6 Cl atoms. For 2 molecules, $2\times6 = 12$.
# Answer:
Kr = 2, Cl = 12

### Problem 31: $5\ce{Al(C_2H_3O_2)_3}$
# Explanation:
## Step1: Calculate Al atoms
In one $\ce{Al(C_2H_3O_2)_3}$, there is 1 Al atom. For 5 molecules, $5\times1 = 5$.
## Step2: Calculate C atoms
In one $\ce{Al(C_2H_3O_2)_3}$, there are $2\times3 = 6$ C atoms. For 5 molecules, $5\times6 = 30$.
## Step3: Calculate H atoms
In one $\ce{Al(C_2H_3O_2)_3}$, there are $3\times3 = 9$ H atoms. For 5 molecules, $5\times9 = 45$.
## Step4: Calculate O atoms
In one $\ce{Al(C_2H_3O_2)_3}$, there are $2\times3 = 6$ O atoms. For 5 molecules, $5\times6 = 30$.
# Answer:
Al = 5, C = 30, H = 45, O = 30

### Problem 32: $3\ce{(NH_4)_2Cr_2O_7}$
# Explanation:
## Step1: Calculate N atoms
In one $\ce{(NH_4)_2Cr_2O_7}$, there are $2$ N atoms. For 3 molecules, $3\times2 = 6$.
## Step2: Calculate H atoms
In one $\ce{(NH_4)_2Cr_2O_7}$, there are $4\times2 = 8$ H atoms. For 3 molecules, $3\times8 = 24$.
## Step3: Calculate Cr atoms
In one $\ce{(NH_4)_2Cr_2O_7}$, there are $2$ Cr atoms. For 3 molecules, $3\times2 = 6$.
## Step4: Calculate O atoms
In one $\ce{(NH_4)_2Cr_2O_7}$, there are $7$ O atoms. For 3 molecules, $3\times7 = 21$.
# Answer:
N = 6, H = 24, Cr = 6, O = 21

### Problem 33: $5\ce{Fe_3(PO_4)_2}$
# Explanation:
## Step1: Calculate Fe atoms
In one $\ce{Fe_3(PO_4)_2}$, there are $3$ Fe atoms. For 5 molecules, $5\times3 = 15$.
## Step2: Calculate P atoms
In one $\ce{Fe_3(PO_4)_2}$, there are $2$ P atoms. For 5 molecules, $5\times2 = 10$.
## Step3: Calculate O atoms
In one $\ce{Fe_3(PO_4)_2}$, there are $4\times2 = 8$ O atoms. For 5 molecules, $5\times8 = 40$.
# Answer:
Fe = 15, P = 10, O = 40

### Problem 34: $2\ce{NH_4NO_3}$
# Explanation:
## Step1: Calculate N atoms
In one $\ce{NH_4NO_3}$, there are $2$ N atoms. For 2 molecules, $2\times2 = 4$.
## Step2: Calculate H atoms
In one $\ce{NH_4NO_3}$, there are $4$ H atoms. For 2 molecules, $2\times4 = 8$.
## Step3: Calculate O atoms
In one $\ce{NH_4NO_3}$, there are $3$ O atoms. For 2 molecules, $2\times3 = 6$.
# Answer:
N = 4, H = 8, O = 6

### Problem 35: $5\ce{BaC_4H_4O_6}$
# Explanation:
## Step1: Calculate Ba atoms
In one $\ce{BaC_4H_4O_6}$, there is $1$ Ba atom. For 5 molecules, $5\times1 = 5$.
## Step2: Calculate C atoms
In one $\ce{BaC_4H_4O_6}$, there are $4$ C atoms. For 5 molecules, $5\times4 = 20$.
## Step3: Calculate H atoms
In one $\ce{BaC_4H_4O_6}$, there are $4$ H atoms. For 5 molecules, $5\times4 = 20$.
## Step4: Calculate O atoms
In one $\ce{BaC_4H_4O_6}$, there are $6$ O atoms. For 5 molecules, $5\times6 = 30$.
# Answer:
Ba = 5, C = 20, H = 20, O = 30

### Problem 36: $4\ce{Cu(HSO_3)_2}$
# Explanation:
## Step1: Calculate Cu atoms
In one $\ce{Cu(HSO_3)_2}$, there is $1$ Cu atom. For 4 molecules, $4\times1 = 4$.
## Step2: Calculate H atoms
In one $\ce{Cu(HSO_3)_2}$, there are $2$ H atoms. For 4 molecules, $4\times2 = 8$.
## Step3: Calculate S atoms
In one $\ce{Cu(HSO_3)_2}$, there are $2$ S atoms. For 4 molecules, $4\times2 = 8$.
## Step4: Calculate O atoms
In one $\ce{Cu(HSO_3)_2}$, there are $3\times2 = 6$ O atoms. For 4 molecules, $4\times6 = 24$.
# Answer:
Cu = 4, H = 8, S = 8, O = 24

### Problem 37: $9\ce{Au(NO_2)_2}$
# Explanation:
## Step1: Calculate Au atoms
In one $\ce{Au(NO_2)_2}$, there is $1$ Au atom. For 9 molecules, $9\times1 = 9$.
## Step2: Calculate N atoms
In one $\ce{Au(NO_2)_2}$, there are $2$ N atoms. For 9 molecules, $9\times2 = 18$.
## Step3: Calculate O atoms
In one $\ce{Au(NO_2)_2}$, there are $2\times2 = 4$ O atoms. For 9 molecules, $9\times4 = 36$.
# Answer:
Au = 9, N = 18, O = 36

### Problem 38: $3\ce{K_2ZnO_2}$
# Explanation:
## Step1: Calculate K atoms
In one $\ce{K_2ZnO_2}$, there are $2$ K atoms. For 3 molecules, $3\times2 = 6$.
## Step2: Calculate Zn atoms
In one $\ce{K_2ZnO_2}$, there is $1$ Zn atom. For 3 molecules, $3\times1 = 3$.
## Step3: Calculate O atoms
In one $\ce{K_2ZnO_2}$, there are $2$ O atoms. For 3 molecules, $3\times2 = 6$.
# Answer:
K = 6, Zn = 3, O = 6

### Problem 39: $3\ce{Sr(MnO_4)_2}$
# Explanation:
## Step1: Calculate Sr atoms
In one $\ce{Sr(MnO_4)_2}$, there is $1$ Sr atom. For 3 molecules, $3\times1 = 3$.
## Step2: Calculate Mn atoms
In one $\ce{Sr(MnO_4)_2}$, there are $2$ Mn atoms. For 3 molecules, $3\times2 = 6$.
## Step3: Calculate O atoms
In one $\ce{Sr(MnO_4)_2}$, there are $4\times2 = 8$ O atoms. For 3 molecules, $3\times8 = 24$.
# Answer:
Sr = 3, Mn = 6, O = 24

### Problem 40: $4\ce{Al_2(CO_3)_3}$
# Explanation:
## Step1: Calculate Al atoms
In one $\ce{Al_2(CO_3)_3}$, there are $2$ Al atoms. For 4 molecules, $4\times2 = 8$.
## Step2: Calculate C atoms
In one $\ce{Al_2(CO_3)_3}$, there are $3$ C atoms. For 4 molecules, $4\times3 = 12$.
## Step3: Calculate O atoms
In one $\ce{Al_2(CO_3)_3}$, there are $3\times3 = 9$ O atoms. For 4 molecules, $4\times9 = 36$.
# Answer:
Al = 8, C = 12, O = 36

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 26: \(12\ce{H_2O}\)

Step1: Calculate H atoms

Step2: Calculate O atoms

Step1: Calculate N atoms

Step2: Calculate O atoms

Step1: Calculate Cl atoms

Step2: Calculate F atoms

Answer:

Problem 27: \(5\ce{N_2O_4}\)