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3 years ago

Calculate the [OH−] value with [H3O+]=3.0×10−9M.

Chemistry

1 answer:

saw5 [17]3 years ago

4 0

Yessssssssdddssssssss

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Compute the following and give your answer on scientific notation

Nata [24]

Answer:

6*10^-3

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Procedure in the attached file

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2 years ago

A 0.100-l sample of an unknown hno3 solution required 32.9 ml of 0.200 m ba(oh)2 for complete neutralization. what was the conce

ddd [48]

The reaction between HNO3 and Ba(OH)2 is given by the equation below;
2HNO3 + Ba(OH)2 = Ba(NO3)2 + 2H2O
Moles of Barium hydroxide used;
= 0.200 × 0.039 l
= 0.0078 Moles
The mole ratio of HNO3 and Ba(OH)2 is 2: 1
Therefore; moles of nitric acid used will be;
= 0.0078 ×2 = 0.0156 moles
But; 0.0156 moles are equal to a volume of 0.10
The concentration of Nitric acid will be;
= (0.0156 × 1)/0.1
= 0.156 M

7 0

3 years ago

Read 2 more answers

The human body is about 9.5% carbon atoms, with a small fraction of that carbon being radioactive carbon-14. Carbon dating is a

maw [93]

Answer:The age can be determined by utilizing carbon-14, a radioactive isotope of carbon.

Explanation:

8 0

2 years ago

What kingdom do fungi belong

miv72 [106K]

They’re Eukaryotic/ the Eukarya kingdom
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8 0

3 years ago

Copper is commonly mined as an ore with a variable percent composition of copper (II) sulfide. This ore is also sometimes referr

pantera1 [17]

Answer:

$m_{CuO}=93.6gCuO$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$CuS+\frac{3}{2} O_2\rightarrow CuO+SO_2$

Thus, given the 1.00-kg of 12.5% ore, we can compute the theoretical yield of copper (II) oxide via stoichiometry:

$m_{CuO}^{theoretical}=1.00kgCuS*\frac{1000gCuS}{1kgCuS} *\frac{12.5gCuS}{100gCuS} *\frac{1molCuS}{95.6gCuS} *\frac{1molCuO}{1molCuS} *\frac{79.5gCuO}{1molCuO} \\\\m_{CuO}^{theoretical}=103.95gCuO$

Whereas the third factor accounts for the percent purity of the covellite. Then, given the percent yield, we can compute the actual yield by:

$m_{CuO}=103.95gCuO*0.9\\\\m_{CuO}=93.6gCuO$

Regards.

6 0

3 years ago