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mote1985 [20]
2 years ago
15

5. When. 0.333g of a particular molecular compound dissolves in 33.3 grams of benzene, the freezing point was observed to drop b

y 0.485 °C. (In actual practice this is absurdly precise for a temperature measurement, but it makes the problem easier to solve.) What is the molecular weight of the molecular compound?
Chemistry
1 answer:
Ivanshal [37]2 years ago
3 0

The molecular weight of the molecular compound is

x= 105.57 g/mol

<h3>What is the molecular weight of the molecular compound?</h3>

Generally,a  the equation for Mole of molecular compound is  mathematically given as

Mm= mass / molar mass

Therefore

Mm= 0.333 g / x

the equation for Morality

M = mole of solute / mass of solvent

M = \frac{(0.333 g / x )}{33.3 g / 1000 }

M= 10 / x

the equation for change in temp dt

dt= K* morality

Therefore

0.485 = \frac{5.12}{ * 10/ x}

x = (5.12 *10) / 0.485

x = 105.57 g/mol

In conclusion, molecular mass of molecular compound

x= 105.57 g/mol

Read more about molecular compound

brainly.com/question/23088724

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bulgar [2K]

Answer:

1.674929 x 10-27 kg

Explanation:

6 0
3 years ago
Give the name of this alcohol.
solong [7]

Answer:

An alcohol is an organic compound with a hydroxyl (OH) functional group on an aliphatic carbon atom. Because OH is the functional group of all alcohols, we often represent alcohols by the general formula ROH, where R is an alkyl group. Alcohols are common in nature.

Explanation:

3 0
3 years ago
How many moles of argon are contained in 58 L of At at STP?
Harman [31]

Answer:

n = 2.58 mol

Explanation:

Given data:

Number of moles of argon = ?

Volume occupy = 58 L

Temperature = 273.15 K

Pressure = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

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

T = temperature in kelvin

1 atm × 58 L = n × 0.0821 atm.L/ mol.K × 273.15 K

58 atm.L = n × 22.43 atm.L/ mol.

n = 58 atm.L / 22.43 atm.L/ mol

n = 2.58 mol

8 0
3 years ago
Question
madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from  22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

5 0
3 years ago
1. A compound is found to contain
satela [25.4K]

Answer:

FeSO2

Explanation:

Please see attached picture for full solution.

8 0
3 years ago
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