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Naddika [18.5K]
3 years ago
14

A solution is prepared by dissolving sugar in water. The solution is 25.0%, by mass, sugar. How many grams of WATER are in 472 g

rams of this solution?
Chemistry
1 answer:
const2013 [10]3 years ago
3 0
If the sugar makes up 25% of the solution, that means the water makes up 75%, or 3/4, of the solution.

Set up a proportion to solve:
\frac{3}{4} = \frac{x}{472}

Cross multiply:
4x = 1416
x = 354

Therefore, there are 354 grams of water in the solution.
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Using the standard enthalpies of formation found in the textbook, determine the enthalpy change for the combustion of ethanol c2
ArbitrLikvidat [17]
Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)

Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol

Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol

<em>Answer: -1367.5 kJ/mol</em>
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3 years ago
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Which sample of water contains particles having the highest kinetic energy
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Answer:It got to be C

Explanation: 75ml of water is

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Which statement best describes the main benefit to organisms that reproduce sexually?
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Answer:

B. They will conserve energy during reproduction

Explanation:

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The atmosphere we live in is about 5-9 miles thick, depending on where you are on Earth. It is a mixture of 78% nitrogen, 21% ox
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Answerino by Duderino:

The troposphere

3 0
3 years ago
A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica
Sladkaya [172]

Answer: The empirical formula for the given compound is CH_5N

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = \frac{3.23}{3.23}=1

For Hydrogen  = \frac{16.2}{3.23}=5.01\approx 5

For Oxygen  = \frac{3.24}{3.23}=1.00\approx 1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is C_1H_5N_1=CH_5N

3 0
3 years ago
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