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3 years ago

This is something easy can someone please finish this one. i'll give brainliest.

Chemistry

2 answers:

Ainat [17]3 years ago

4 0

Answer:

B B C A C B A A C C

Explanation:

jk it was a big guess

Flauer [41]3 years ago

4 0

Answer:

1. C maybe

2. C

3. B

4. A

5. A

6. B

7. C

8. A

9. C

10. A

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What is Covalent bond?

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4 years ago

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A solution contains 0.775 g Ca2 0.775 g Ca2 in enough water to make a 1925 mL1925 mL solution. What is the milliequivalents of C

Romashka [77]

Answer: There will be 0.00002 meq per Liter of the solution.

Explanation:

Normality is defined as the umber of gram equivalents dissolved per liter of the solution.

$Normality=\frac{\text{no of gram equivalents} \times 1000}{\text{Volume in ml}}$

$Normality=\frac{\text{given mass}\times 1000}{\text {Equivalent mass}\times {\text{Volume in ml}}}$

Equivalent weight is calculated by dividing the molecular weight by n factor. ${\text{Equivalent weight}}=\frac{\text{Molecular weight}}{n}$

n= charge for charged species , For $Ca^{2+}$ , n =2

${\text{Equivalent weight}}=\frac{40}{2}=20g$

$\text{no of gram equivalents}=\frac{0.775g}{20g/mol}=0.04gramequivalents$

$Normality=\frac{0.04\times 1000}{1925}=0.02eq/L=0.00002meq/L$

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4 years ago

Which combination may be used to prepare a buffer having a ph of 8. 8?

7nadin3 [17]

The combination used in the preparation of a buffer with a pH around 8.8 accounts for ka = 7 * 10 -3 for h 3po4

The ph of the buffer can be shown as:

pH = pKa + log [Salt] /[ Acid ]

[Salt] /[ Acid ] = x

For h3po4 with ka= 7 × 10–3

8.8 = - log (7 × 10^–3) + log x

8.8 = 2.21 + log x

Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

For h2po4- with ka= 8 × 10–8

8.8 = - log (8 × 10^–8) + log x

8.8 = 7.14 + log x

Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

For hpo42– with ka= 5 × 10–13

8.8 = - log (5 × 10–13) + log x

8.8 = 12.31 + log x

Thus, the value of log x is coming negative and therefore can not be used for preparing buffer.

Hence, the correct answer is option A

Learn more about buffering systems here,

brainly.com/question/16556401

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2 years ago

A compound contains only carbon, hydrogen, and oxygen. combustion of 11.75 mg of the compound yields 17.61 mg co2 and 4.81 mg h2

Ymorist [56]

Number of moles is defined as the ratio of given mass in g to the molar mass.

First, convert the given mass of carbon dioxide in mg to g:

1 mg = 0.001 g

17.61 mg = 0.01761 g

Number of moles of carbon dioxide = $\frac{0.01761 g}{44.01 g/mol}$

= $0.0004001 mol$

Mass of carbon = number of moles of carbon dioxide \times molar mass of carbon

= $0.0004001 mol\times 12.011 g/mol$

= $0.004806 g$

Number of moles of water= $\frac{0.00481 g}{18 g/mol}$

= $2.672\times 10^{-4}$

Since, water contains two hydrogen atoms. Thus,

Moles of hydrogen = $2\times 2.672\times 10^{-4}$

= $5.34\times 10^{-4}$

Mass of hydrogen = $5.34\times 10^{-4}\times \times 1.008 g/mol$

= $5.34\times 10^{-4} g$

Mass of oxygen = $0.001175-(5.38\times 10^{-4}g+0.004806 g)$

= $0.006405 g$

Number of moles of oxygen = $\frac{0.006405 g}{15.999 g/mol}$

= $0.000400$

Now,

$C_{0.0004001} H_{0.000534} O_{0.000400}$

Divide the smallest number to get the whole number,

$C_{\frac{0.0004001}{0.000400}} H_{\frac{0.000534}{0.000400}} O_{\frac{0.000400}{0.000400}}$

we get,

$C_{1} H_{1.33} O_{1}$

Now, multiply all the subscript by 3 to get the whole number,

$C_{3} H_{4} O_{3}$ (empirical fomula)

Molar mass of the compound = $3\times 12.011 g/mol+4\times 1.008 g/mol+3\times 15.999 g/mol$

= $88.062 g/mol$

Divide given molar mass of the compound with the molar mass of the compound.

= $\frac{176.1 g/mol}{88.062 g/mol}$

= $1.999\simeq 2$

Thus, multiply the subscripts of empirical formula by 2 to get the molecular formula, we get:

$C_{6}H_{8}O_{6}$

Hence, empirical formula is $C_{3}H_{4}O_{3}$ and molecular formula is $C_{6}H_{8}O_{6}$

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