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3 years ago

Can someone help me with this?

Chemistry

1 answer:

Nadusha1986 [10]3 years ago

4 0

Answer:

idk my man srry5678976543567890876546789087659687876875787678687367376367373

Explanation:

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A potassium ion (K+) would most likely bond with?

ivanzaharov [21]

It's most likely to bond with chlorine.

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3 years ago

Calcula la masa equivalente del clorato de calcio CA (CI03)2

Vitek1552 [10]

Answer:

207g

Explanation:

Given compound:

Ca(ClO₃)₂

Equivalent mass is the

The equivalent mass can be derived by summing the molar masses of each atom

Molar mass of Ca = 40

Cl = 35.5

O = 16

Now solve;

Molar mass = 40 + 2(35.5 + 3(16)) = 207g

6 0

3 years ago

How many grams of H2O are produced when 16.3g of O2

seraphim [82]

32.7 grams will be produced

7 0

3 years ago

Please help me!!

docker41 [41]

I also think it C too

8 0

3 years ago

Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a cat

Kobotan [32]

<u>Answer:</u> The percent yield of the reaction is 32.04 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 112.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{112.12g/mol}=0.0321mol$

The chemical equation for the formation of methyl benzoate from benzoic acid follows:

$\text{Benzoic acid + Methanol}\rightarrow \text{Methyl benozate}$

By Stoichiometry of the reaction:

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0321 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0321=0.0321mol$ of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0321 moles

Putting values in equation 1, we get:

$0.0321mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0321mol\times 136.15g/mol)=4.37g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.4 g

Theoretical yield of methyl benzoate = 4.37 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.4g}{4.37g}\times 100\\\\\% \text{yield of methyl benzoate}=32.04\%$

Hence, the percent yield of the reaction is 32.04 %.

6 0

3 years ago