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Iteru [2.4K]
3 years ago
6

Question 2 (2 points)

Chemistry
1 answer:
poizon [28]3 years ago
7 0

Answer:

m_w=439.2g

Explanation:

Hello!

In this case, since the by-mass percent of a solution is a measure of the mass of the solute over the mass of the solution:

\%m/m=\frac{m_{solute}}{m_{solution}} *100\%

As we know the mass of the solution and the by-mass percent, we can compute the mass of glucose in the 480 g of solution:

m_{solute}=\frac{\%m/m*m_{solution}}{100\%}

Thus, by plugging in the data, we obtain:

m_{solute}=\frac{8.5\%*480g}{100\%}=40.8g

Finally, since the solution is made up of glucose and water, we compute the mass of water as follows:

m_w=m_{sol}-m_{solute}=480g-40.8g\\\\m_w=439.2g

Best regards!

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The elemental mass percent composition of ascorbic acid (vitamin C ) is 40.92% C , 4.58% H , and 54.50% O . Determine the empiri
Lady bird [3.3K]

The empirical formula of ascorbic acid is C6H8O6.

<h3>Empirical formula</h3>

To calculate the empirical mass of a compound from the mass percentage of each element, the value of the<u> molar mass </u>of each element is used, which in this case corresponds to:

                                              MM_C= 12g/mol\\MM_H=1g/mol\\MM_O=16g/mol

From this, we consider that the compound has 100 grams, so the number of moles of each element will be equal to:

                                                   C = \frac{40.92}{12} = 3.41 \\H = \frac{4.58}{1}=4.58\\ O =\frac{54.50}{16}=3.41

Now, divide all the values ​​found by the <u>smallest value</u>, to find the amount present in each element:

                                            C = \frac{3.41}{3.41} = 1\\ H =  \frac{4.58}{3.41} = 1.34\\O =  \frac{3.41}{3.41} = 1

As you can see, the value of moles of hydrogen resulted in a decimal number, so it is necessary to multiply all values ​​​​by a number in common until the three meet as integers:

                                          C = 1 \times 6 = 6\\H = 1.34 \times 6 = 8\\O = 6 \times 6 = 6

So, the empirical formula of ascorbic acid is C6H8O6.

Learn more about empirical and molecular formula in: brainly.com/question/11588623

7 0
3 years ago
Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2
Svet_ta [14]

Answer:

-1815.4 kJ/mol

Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)

This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

In this case:

ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol

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