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2 years ago

How many moles is 130g of cocl2

Chemistry

1 answer:

Oduvanchick [21]2 years ago

8 0

Answer:

1.3142 mol CoCl2

Explanation:

Well, to answer this we need to find out how many grams are in a single mole of CoCl2. Two find that we need to get the molar mass of CoCl2

Co = 28.0101 g Co per mole

Cl = 35.453 * g Cl per mole

28.0101 + 2 * 35.453 = 98.9161 g CoCl2 per mole

Now, we divide:

130/98.9161 = 1.3142 mol CoCl2

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One liter of ocean water contains 35.06 g of salt. What volume of ocean water would contain 1.00 kg of salt? Express your answer

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Answer:

28.52 L

Explanation:

First, let's calculate the density of the ocean, which is the mass divided by the volume:

d = m/V

d = 35.06/1

d = 35.06 g/L

So, for a mass of 1.00 kg = 1000.00 g

d = m/V

35.06 = 1000.00/V

V = 1000.00/35.06

V = 28.52 L

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2 years ago

the melting point of scandium fluoride is 1552°c, explain why scandium fluoride has a high melting point.

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Answer:

Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.

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2 years ago

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Complete this equation for the dissociation of the electrolyte NaCl(aq). Omit water from the equation because it is understood t

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1. Answer;

Equation;

-NaCl(aq) → Na{+}(aq) + Cl{-}(aq)

Explanation;

-The polar property of water allows ionic compounds such as sodium chloride to dissolve in water. (sodium chloride is an ionic compound).

-Sodium chloride dissociates in water into sodium ions and chloride ions as shown by the equation. Water molecules surrounds the negatively charged chloride ions and positive sodium ions and pulls them away into solution.

2. Answer;

Equation;

Fe(ClO4)3(aq) → Fe 3+(aq) + 3ClO4 - (aq)

Explanation;

Iron (iii) perchlorate (Fe(ClO4)3 dissociates in water into iron (iii) ion and perchlorate ion. Iron (iii) ion is positively charged while the perchlorate ion is negatively charged, they are pulled apart by the polar water molecules, a process called dissociation.

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2 years ago

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An empty beaker is weighed and found to weigh 23.1 g. Some potassium chloride is then added to the beaker and weighed again. The

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Answer:Mass of Potassium chloride =1.762g

Explanation:

Mass of empty beaker = 23.100 g

Mass of beaker with Potassium chloride = 24.862g

Mass of Potassium chloride = Final weight - initial weight = Mass of beaker with Potassium chloride - Mass of empty beaker = 24.862-23.100 = 1.762g

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2 years ago

A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture

Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

$n=\frac{PV}{RT}$

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

$n=(\frac{1.45*2.50}{0.0821*293})$

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide = $8.60g(\frac{1mol}{44g})$

moles of carbon dioxide = 0.195 mol

Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

0.195 = 0.453 - 2X

2X = 0.453 - 0.195

2X = 0.258

$X=\frac{0.258}{2}$

X = 0.129

So, there are 0.129 moles of methane in the mixture.

moles of propane = 0.151 - 0.129 = 0.022

mole fraction of methane = $\frac{moles of Methane}{total moles}$

mole fraction of methane = $\frac{0.129}{0.151}$

mole fraction of methane = 0.854

Hence, the mole fraction of methane gas in the original gas mixture is 0.854.

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2 years ago