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Nutka1998 [239]
2 years ago
12

Ocean life is changing due toThis is happening becauseThis is causingbecause​

Chemistry
1 answer:
nadya68 [22]2 years ago
8 0

Answer:

what can you give more details of what this is i would like to help

Explanation:

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The poh of a solution is 9.32. calculate the hydrogen ion concentration of the solution.
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PH+pOH=14
pOH=9.32
pH=14-pOH=14-9.32=4.68
pH=4.68

pH=-log[H⁺]
-pH= log[H⁺]

[H^{+}]= 10^{-pH} = 10^{-4.68}=2.09*10^{-5}

[H^{+}] = 2.09*10^{-5}M
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3 years ago
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)a. Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of
Ira Lisetskai [31]

Answer:

a. 167 mL.

b. 39.3 %.

Explanation:

Hello!

In this case, for the undergoing chemical reaction, since 45.0 g of aluminum react, based on the 2:3 mole ratio with sulfuric acid, we can compute the required moles  as shown below:

n_{H_2SO_4}=45.0gAl*\frac{1molAl}{27.0gAl} *\frac{3molH_2SO_4}{2molAl} =2.50molH_2SO_4

Next, since the molarity of a solution is computed based on the moles and volume (M=n/V), we can compute the required volume of sulfuric acid as shown below:

V=\frac{n}{M}=\frac{2.50mol}{15.0mol/L}=0.167L

That in mL is 167 mL.

Moreover, for the percent yield, we compute the grams of aluminum sulfate that are produced based on the required 2.50 moles of sulfuric acid:

m_{Al_2(SO_4)_3}=2.50molH_2SO_4*\frac{1molAl_2(SO_4)_3}{3molH_2SO_4}*\frac{342.15gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=285.13gAl_2(SO_4)_3

Therefore the percent yield is:

Y=\frac{112g}{285.13g}*100\%\\\\Y=39.3\%

Best regards!

6 0
2 years ago
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