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klio [65]
3 years ago
5

Calculate the mass of sucrose necessary to make a 5% by mass sucrose solution if the solution contains 50.0 ml of distilled wate

r. (Hint: the total mass of the solution is the mass of sucrose plus the mass of water.)
Chemistry
1 answer:
Lelechka [254]3 years ago
5 0

Answer:

2.6 g

Explanation:

Percent concentration = 5%

Volume of water = 50ml

Since 1ml of water = 1g of water

Mass of water= 50g

Let the mass of sucrose be x

Mass% = mass of solute/mass of solute + mass of solvent × 100

5= x/x+ 50 ×100

5/100 = x/x + 50

0.05 = x/x + 50

0.05(x + 50) = x

0.05x + 2.5 = x

2.5 = x - 0.05x

2.5 = 0.95x

x= 2.5/0.95

x=2.6 g

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How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.88 ∘c?
e-lub [12.9K]
The freezing point depression is calculated through the equation,
                                    ΔT = (kf)  x m 
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
                                   5.88 = (1.86)(m)
m is equal to 3.16m

Recall that molality is calculated through the equation,
                                  molality = number of mols / kg of solvent
                                       number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.

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3 years ago
What is it??!!!!!??!??!?
Softa [21]

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other the effect of temperature on plants

4 0
3 years ago
Flammability is a material’s ability to burn in the presence of
frosja888 [35]

Answer:

Flammability is a material’s ability to burn in the presence of <u><em>oxygen.</em></u>

Explanation:

Flammability can be described as the ability of a substance to get ignited. Flammability will lead to fire or combustion. Some substances are highly flammable like Benzene. Other tend to be just flammable. And there are also compounds which will nor be flammable at all as they won't react with oxygen. Examples of these substances include helium, steel or glass.

The flammability of a substance shall be considered a very important aspect when storing or transporting a substance.

5 0
3 years ago
Aqueous solutions of isopropyl alcohol are commonly sold as rubbing alcohol. The boiling point of isopropyl alcohol is 82.4 °C.
Monica [59]

Answer:

This is due to more hydrogen bonding in ethylene glycol than it is in isopropyl alcohol

Explanation:

The boiling point of isopropyl alcohol is 82.4 °C it contains only a single OH group, hence intermolecular hydrogen bonding is solely responsible for it's boiling point, whereas Ethylene glycol (CH2OHCH2OH) contains 2-OH group and both intermolecular and intramolecular hydrogen bonding are responsible for the higher boiling point of ethylene glycol at 198 °C.

8 0
3 years ago
When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
algol [13]

Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

Mass of H_2O= 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, =\frac{2}{18}\times 7.759=0.862g of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

6 0
3 years ago
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