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3 years ago

Which one of the following substances would be the most soluble in CCl4?

Chemistry

1 answer:

maria [59]3 years ago

5 0

Answer: Option (d) is the correct answer.

Explanation:

Carbon tetrachloride $(CCl_{4})$ is a non-polar solvent. Whereas out of the given options, $Na_{2}SO_{4}$ , $H_{2}O$ , $CH_{3}CH_{2}CH_{2}CH_{2}OH$ , and HI are all polar molecules.

On the other hand, only $C_{4}H_{10}$ is non-polar molecule.

Also it is known that like dissolves like.

So, being non-polar $CCl_{4}$ will dissolve the give alkane, $C_{4}H_{10}$ .

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Slav-nsk [51]

Temperature of somewhere.

6 0

3 years ago

Complete the following radioactive decay problem. 222 86 RN to 4 2 HE + ?

Andreas93 [3]

Answer:

₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴

Explanation:

The given nuclear reaction shows alpha decay.

₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴

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Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number less than 2 as compared to parent atom the starting atom.

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₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + energy

8 0

3 years ago

Potassium chlorate is a white, crystalline solid. It can be broken down into potassium chloride (a salt)

lions [1.4K]

Answer:

Notice that the number of atoms of

and

are the same on both sides, but the numbers of

atoms are not. There are 3

atoms on the the left side and 2 on the right. 3 and 2 are factors of 6, so add coefficients so that there are 6

atoms on both sides.

2KClO

(

)

+ heat

→

KCl(s)

(

)

Now the

and

atoms are not balanced. There are 2 of each on the left and 1 of each on the right. Add a coefficient of 2 in front of

KCl

2KClO

(

)

+ heat

→

2KCl(s)

(

)

The equation is now balanced with 2

atoms,

7 0

3 years ago

Read 2 more answers

An iron block of mass 18 kg is heated from 285 K to 318 K. If 267.3 kJ is required, what is the specific heat of iron? A. 450.00

valkas [14]

Answer:

Option A. 450.00

Explanation:

1) Data:

a) m = 18 kg

b) T₁ = 285 K

c) T₂ = 318 K

d) Q = 267.3 kJ

e) S = ?

2) Principles and equations

The specific heat of a substance is the amount of heat energy absorbed to increase the temperature of certain amount (gram, kg, or moles, depending on the definition or units) of the substance in 1 ° C or 1 K.

The mathematical relation between the specific heat and the heat energy absorbed is:

Q = m × S × ΔT

Where,

Q is the heat absorbed,
S is the specific heat, and
ΔT is the temperature increase (T₂ - T₁)

3) Solution:

a) Substitute the data into the equation:

267.3 kJ = 18 kg × S × (318 K - 285 K)

b) Solve for S and compute:

S = 267.3 kJ / (18 kg × 33 K) = 0.45 kJ / (Kg . K)

The options have not units, but I notice that the first answer is 1,000 times the answer I obtained, so I will make a conversion of units.

c) Convert to J /( kg . k):

0.45 kJ / (Kg . K) × 1,000 J / kJ = 450 J / (kg . K)

Now we can see that the option A is is the answer, assuming the units.

6 0

3 years ago

A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and

Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = $99.7^oC$

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

$1atm=760mmHg$

or,

$1mmHg=\frac{1}{760}atm$

As, $1mmHg=\frac{1}{760}atm$

So, $752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm$

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

$K=273+^oC$

As, $K=273+^oC$

So, $K=273+99.7=372.7$

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor = 0.2481 L

n = number of moles of vapor = ?

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

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

$(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole$

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

$\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}$

$\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole$

Thus, the mass of one mole of vapor is 32.0 g/mole

8 0

3 years ago