fill in the boxes for the coefficient and subscripts to make the formula match the number of atoms. the first one has been done for you as an example.
example: kr = 2 cl = 12 2 krcl₆
3) k = 6 zn = 3 o = 6 3 k₂zno₂
4) ba = 5 c = 20 h = 20 o = 30 □ bac□h□o□
5) cu = 4 h = 8 s = 8 o = 24 □ cu(hso□)□
6) n = 6 h = 24 cr = 6 o = 21 □ (nh□)□ cr□o□

Question

Accepted Answer

# Explanation:
## Step1: Analyze the formula $BaC_{x}H_{y}O_{z}$ for question 4
We know that the coefficient $a$ and sub - scripts $x,y,z$ should satisfy the atom - count equations. For $Ba$: $a\times1 = 5$ (since there is 1 $Ba$ atom in $BaC_{x}H_{y}O_{z}$), so $a = 5$. For $C$: $a\times x=20$, substituting $a = 5$, we get $5x = 20$, then $x=\frac{20}{5}=4$. For $H$: $a\times y = 20$, substituting $a = 5$, we get $5y=20$, then $y = 4$. For $O$: $a\times z=30$, substituting $a = 5$, we get $5z = 30$, then $z = 6$.
## Step2: Analyze the formula $Cu(HSO_{x})_{y}$ for question 5
For $Cu$: $a\times1=4$ (since there is 1 $Cu$ atom in $Cu(HSO_{x})_{y}$), so $a = 4$. For $H$: $a\times y\times1=8$ (1 $H$ atom in $HSO_{x}$), substituting $a = 4$, we get $4y=8$, then $y = 2$. For $S$: $a\times y\times1=8$ (1 $S$ atom in $HSO_{x}$), substituting $a = 4$ and $y = 2$ (already found from $H$ - count), it is consistent. For $O$: $a\times y\times x=24$, substituting $a = 4$ and $y = 2$, we get $4\times2\times x=24$, then $8x=24$, $x = 3$.
## Step3: Analyze the formula $(NH_{x})_{y}Cr_{z}O_{w}$ for question 6
For $N$: $a\times y\times1=6$ (1 $N$ atom in $NH_{x}$). For $H$: $a\times y\times x=24$. For $Cr$: $a\times z\times1=6$ (1 $Cr$ atom). For $O$: $a\times w\times1=21$. Let's assume $a = 6$ (from $Cr$ - count: if $z = 1$, then $a=6$). For $N$: if $a = 6$ and $y = 1$ (since $a\times y\times1=6$), it works. For $H$: $6\times1\times x=24$, then $x = 4$. For $O$: if $a = 6$ and $w=\frac{21}{6}=\frac{7}{2}$ (not valid). Let's try another approach.
Let $a = 3$. For $Cr$: $a\times z=6$, then $z = 2$. For $N$: $a\times y=6$, if $y = 2$. For $H$: $a\times y\times x=24$, substituting $a = 3$ and $y = 2$, we get $3\times2\times x=24$, then $x = 4$. For $O$: $a\times w=21$, substituting $a = 3$, we get $w = 7$.

# Answer:
4) $\boxed{5}BaC_{\boxed{4}}H_{\boxed{4}}O_{\boxed{6}}$
5) $\boxed{4}Cu(HSO_{\boxed{3}})_{\boxed{2}}$
6) $\boxed{3}(NH_{\boxed{4}})_{\boxed{2}}Cr_{\boxed{2}}O_{\boxed{7}}$

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the formula \(BaC_{x}H_{y}O_{z}\) for question 4

Step2: Analyze the formula \(Cu(HSO_{x})_{y}\) for question 5

Step3: Analyze the formula \((NH_{x})_{y}Cr_{z}O_{w}\) for question 6

Answer: