2 years ago

How many elements in CuSO4

Chemistry

2 answers:

belka [17]2 years ago

6 0

Three. Copper (Cu), Sulfur (S), and Oxygen (O)

babymother [125]2 years ago

3 0

Cu, S, and O so three.

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Dmitry_Shevchenko [17]

Answer:

1. $Rate =k [NO]^{2}[Cl_{2}]$

2. $k= 0.42 \frac{L^{2}}{mol^{2}*s}$

Explanation:

$Rate =k [NO]^{m}[Cl_{2}]^{n}$

$Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2$

$Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1$

$Rate =k [NO]^{2}[Cl_{2}]^{1}$

$Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}$