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SashulF [63]
1 year ago
10

48. Which of the following (a) KCI (b) NaCl is called Sylvine? (c) CaCl₂ (d) MgCl₂​

Chemistry
1 answer:
White raven [17]1 year ago
5 0

Answer:

(a) KCI

sylvine, is potassium chloride (KCl) in natural mineral form

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How many grams of hydrogen are contained in 2.00 mol of C6H7N
Setler [38]

Answer:- 14.0 moles of hydrogen present in 2.00 moles of [tex]C_6H_7N .

Solution:- We have been given with 2.00 moles of C_6H_7N and asked to calculate the grams of hydrogen present in it. It's a two step conversion problem. In first step we convert the moles of the compound to moles of hydrogen as one mol of the compound contains 7 moles of hydrogen. In next step the moles are converted to grams on multiplying the moles by atomic mass of H. The calculations are shown as:

2.00molC_6H_7N(\frac{7molH}{1molC_6H_7N})(\frac{1.0gH}{1molH})

= 14.0 g H

So, there are 14.0 g of hydrogen in 2.00 moles of  C_6H_7N .

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3 years ago
How can a scientist prevent bias in a scientific investigation
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They must make sure that all things pointed out are fact, not opinion.
5 0
2 years ago
How do substances and mixtures diifer
vlabodo [156]

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5 0
3 years ago
Property of a substance
Scrat [10]

There are many properties to substances.

I'll list some examples below:

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8 0
2 years ago
Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

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

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0
2 years ago
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