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erica [24]
3 years ago
7

A 20.0 ml sample of glycerol has a mass of 25.2 grams. What is the mass of 63 ml sample of glycerol?

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

The mass of 63 ml sample : 79.38 g

<h3>Further explanation</h3>

Given

20 ml and 25.2 g of glycerol

Required

The mass of 63 ml sample

Solution

Density is the ratio of mass per unit volume

Density formula:

\large{\boxed {\bold {\rho~=~ \frac {m} {V}}}}

Density of glycerol :

= m : V

= 25.2 g : 20 ml

= 1.26 g/ml

Mass of 63 ml sample :

= density x volume

= 1.26 g/ml x 63 ml

= 79.38 g

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How are the cells produced through mitosis and meiosis different?
Sholpan [36]

Answer:

In mitosis 2 identical cells are created, but in meiosis 4 sex cells are created.

Explanation:

In mitosis there are 5 stages. These stages are prophase, prometaphase, metaphase, anaphases, and telophase. In meiosis there are also 5 stages, they are the same phases, they just end with II. (Ex. Prophases II)

8 0
3 years ago
. KMnO4 oxidizes ethyl benzene to benzoic acid. If this reaction generally results in a 40 % experimental yield, how many grams
lyudmila [28]

Answer:

530.835 g

Explanation:

First we convert 244 g of benzoic acid (C₇H₆O₂) to moles, using its molar mass:

  • 244 g benzoic acid ÷ 122.12 g/mol = 2.00 moles benzoic acid

Theoretically,<em> one mol of ethyl benzene would produce one mol of benzoic acid</em>. But the experimental yield tells us that one mol of ethyl benzene will produce only 0.4 moles of benzoic acid.

With the above information in mind we convert 2.00 moles of benzoic acid into moles of ethyl benzene:

  • 2.00 moles benzoic acid * \frac{1molEthylBenzene}{0.4molBenzoicAcid} = 5.00 moles ethyl benzene

Finally we <u>convert moles of ethyl benzene </u>(C₈H₁₀)<u> into grams</u>, using its <em>molar mass</em>:

  • 5.00 moles ethyl benzene * 106.167 g/mol = 530.835 g ethyl benzene
7 0
3 years ago
When you add water to calcium oxide it makes calcium hydroxide
Romashka [77]
Yes you add water to calcium oxide it makes calcium hydroxide
7 0
4 years ago
B. Flourine is the right answer
ad-work [718]

Answer:

Whts ur question??

..?

Explanation:

Was this by mistake or smthin

7 0
2 years ago
The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M
anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0
3 years ago
Read 2 more answers
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