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3 years ago

How many grams of NaCl are needed to prepare 1.20 liters of a 2.00 M solution?

Chemistry

1 answer:

larisa86 [58]3 years ago

3 0

Answer: D) 140g

Explanation: no. of moles of NaCl = molarity X volume in litres = 2 X 1.2 = 2.4, and molar mass or mass of 1 mole of NaCl = 58.44 g, so 2.4 moles NaCl = 140.256 g

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Which metal atom below would Not be involved in formation of a Type II Compound?

BartSMP [9]

In formation of a Type II Binary Compound, the metal atom present is<span> NOT</span> found in either Group 1 or Group 2 on the periodic table. For the choices, Ba is under Group 2 on the periodic table, which makes it the atom not involved in formation of type II compounds. The answer is B.

4 0

3 years ago

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0

3 years ago

In chemistry, what varies with the number of molecules present in a sample of a particular substance?

Neko [114]

Answer: concentration

Explanation:

Concentration refers to the amount of a substance present in a sample. The more molecules of a substance present in a sample, the greater its concentration. The less molecules of a substance in a sample, the lesser the concentration. We are often concerned about analytically determining the concentration of a substance using diverse analytical methods in chemistry.

3 0

3 years ago

How many moles are in 17 grams of magnesium chloride (MgCl2) ?

seraphim [82]

Answer:

Molar mass = (24.31 + 2 × 35.45) = 95.21 g mol–1 i.e. 95.21 g of MgCl2 is exactly 1 mole.

Explanation:

7 0

3 years ago

How to convert 5.2 km to mm

ZanzabumX [31]

5.2 km to mm.
1km = 1000m
= 5.2 * 1000m.
=5200 m. 1m = 1000mm

= 5200 * 1000mm
= 5200000 mm

4 0

3 years ago