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3 years ago

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A student carefully pipets 10.00ml of ethanol into a beaker. If ethanol has a density of 0.779g/ml, how many grams of ethanol we

re put into the beaker?

1 answer:

lesantik [10]3 years ago

5 0

Answer: 7.79 grams of ethanol were put into the beaker.

Explanation:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.779 g/mL

Volume of water = 10.00 mL

Putting values in above equation, we get:

$0.779g/mL=\frac{\text{Mass of ethanol}}{10.00mL}\\\\\text{Mass of ethanol}=(0.779g/mL\times 10.00mL)=7.79g$

Thus 7.79 grams of ethanol were put into the beaker.

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Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
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Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

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3 years ago

Aspirin sun thesis Green Chemistry and Assime the aspirin is prepared by the following reaction and that 10.09. of salicylic aci

klemol [59]

<u>Answer:</u> The percentage yield of aspirin is 38.02 %.

<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)

<u>For salicylic acid:</u>

Given mass of salicylic acid $(C_7H_6O_3)$ = 10.09 g

Molar mass of salicylic acid $(C_7H_6O_3)$ = 138.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of salicylic acid}=\frac{10.09g}{138.12g/mol}=0.0730mol$

The chemical equation for the formation of aspirin follows:

$C_7H_6O_3+C_4H_6O_3\rightarrow C_9H_8O_4+CH_3COOH$

As, acetic anhydride is present in excess. So, it is considered as an excess reagent.

Thus, salicylic acid is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of salicylic acid produces 1 mole of aspirin.

So, 0.0730 moles of salicylic acid will produce = $\frac{1}{1}\times 0.0730=0.0730mol$ of aspirin

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

Molar mass of aspirin = 180.16 g/mol

Moles of aspirin = 0.073 moles

Putting values in equation 1, we get:

$0.073mol=\frac{\text{Mass of aspirin}}{180.16g/mol}\\\\\text{Mass of aspirin}=13.15g$

To calculate the percentage yield of aspirin, we use the equation:

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

Experimental yield of aspirin = 5.0 g

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Putting values in above equation, we get:

$\%\text{ yield of aspirin}=\frac{5.0g}{13.15g}\times 100\\\\\% \text{yield of aspirin}=38.02\%$

Hence, the percent yield of aspirin is 38.01 %.

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3 years ago

A 20.00-mL sample of a weak base is titrated with 0.0568 M HCl. At the endpoint, it is found that 17.88 mL of titrant was used.

nata0808 [166]

Answer: 0.0508mL

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therefore concentration of the base is 1.0156/20 = 0.0508 mL

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2 years ago