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4 years ago

You have 535 g of Zinc Chloride, ZnCl2. How many moles is this?

Chemistry

2 answers:

podryga [215]4 years ago

8 0

Answer:3.93moles

Explanation:

I mole of ZnCl2= 65+2(35.5)=136

If 136g of Zinc chloride,,=136grams

Then 535g is 535*1/136=3.93 moles of zinc chloride.

Noted that one mole of a substance contains its molar mass.

makkiz [27]4 years ago

5 0

Best Answer: 1 mole of zinc has a mass of 65.4 grams

Therefore, if 1 mole is 65.4 grams, how many moles is 0.535g?

1 mol = 65.4g

? mol = 0.535g

0.535g

---------

65.4g/mol

(0.535g divided by 65.4grams per mole of Zinc)

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Rank the following substances in order from most soluble in water to least soluble in water: methane, CH4; 2-butanol, C4H9OH; po

nata0808 [166]

Answer:

Arranged in order of most soluble to least soluble in water:

$KCl > CH_{4} > C_{4} H_{9} OH > C_{3}H_{8}$

Explanation:

Water is a polar solvent due to the electric charge it possesses. Hence it would only dissolve polar molecules.

Non-polar molecules would be dissolved by non-polar solvents.

Short chain alcohols are more soluble than butanol, but generally the polar -OH group is soluble in water.

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3 years ago

Ocean water contains 3.5 % NaCl by mass. What mass of ocean water in grams contains 46.8 g of NaCl?

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Answer:

1.3 kg

Explanation:

Mass of ocean water = mass of NaCl × (mass of ocean water/mass NaCl).

The % NaCl tells us there are 3.5 g NaCl/100 g ocean water or 100 g ocean water/3.5 g NaCl.

Mass of ocean water = 46.8 × (100/3.5)

Mass of ocean water = 46.8 × 28.6

Mass of ocean water = 1300 g = 1.3 kg

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4 years ago

Manganese (Mn) has many possible oxidation numbers (+2 and +4 are just two of these). Write the correct formulas and names for t

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Answer:

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2. Potassium permanganate

Its chemical formula is $KMnO_4$ . It is commonly used in the laboratories as oxidizing agent. They appear black in color.

3 Manganese bromide

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4. Oxides of manganese

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3 years ago

Which layer of the earth's atmosphere, is furthest away from the surface of the earth? A) exosphere B) mesosphere C) stratospher

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A) exosphere

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4 years ago

Read 2 more answers

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago