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patriot [66]
3 years ago
13

Calculate the number of moles in 9 g of water​

Chemistry
2 answers:
Slav-nsk [51]3 years ago
8 0

Answer:

0.5 mole

Explanation:

m(h2o)=2×1+16=18g

n=m÷Me=9÷18=0.5mole

tangare [24]3 years ago
7 0

Answer:

= 0.5 moles of water

formular \: mass \: of \: water \:  = (2 \times 1)  + 16  = 18 \\ 18g \: of \: water \: are \: weighed \: by \: 1 \: mole. \\ 9g \: will \: be \: weighed \: by \:  \frac{9 \times 1}{18}  \\  =  \frac{9}{18} moles \\  = 0.5 \: moles \: of \: water

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8 0
2 years ago
if kept at a constant temperature, what is the concentration at 865s if the initial concentration was 0.0225M
Anestetic [448]

Answer:

[SO_2Cl_2]=0.0175M

Explanation:

Hello!

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r=-k[SO_2Cl_2]

We also consider that the integrated rate law has been already reported as:

[SO_2Cl_2]=[SO_2Cl_2]_0exp(-kt)

Thus, by plugging in the initial concentration, rate constant and elapsed time we obtain:

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8 0
2 years ago
A saturated solution of Pb(IO3)2 in pure water has a lead ion concentration of 5.0 x 10-5 Molar. What is the Ksp value of Pb(IO3
Orlov [11]

Answer:

Option (E) is correct

Explanation:

Solubility equilibrium of Pb(IO_{3})_{2} is given as follows-

                   Pb(IO_{3})_{2}\rightleftharpoons Pb^{2+}+2IO_{3}^{-}

Hence, if solubility of Pb(IO_{3})_{2} is S (M) then-

                             [Pb^{2+}]=S(M) and [IO_{3}^{-}]=2S(M)

Where species under third bracket represent equilibrium concentrations

So, solubility product of Pb(IO_{3})_{2} , K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}

Here, [Pb^{2+}]=S(M)=5.0\times 10^{-5}M

So, [IO_{3}^{-}]=2S(M)=(2\times 5.0\times 10^{-5})M=1.0\times 10^{-4}M

So, K_{sp}=(5.0\times 10^{-5})\times (1.0\times 10^{-4})^{2}=5.0\times 10^{-13}

Hence option (E) is correct

7 0
3 years ago
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