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LiRa [457]
3 years ago
6

A 50.0 mL graduated cylinder has a mass of 65.1 g. When it is filled with an unknown liquid to the 49.3 mL mark, the cylinder an

d liquid combined have a mass of 120.5 g. Calculate the density of the liquid in g/cm3.
Chemistry
1 answer:
Delvig [45]3 years ago
3 0

<u>Answer:</u> The density of liquid is 1.12g/cm^3

<u>Explanation:</u>

We are given:

Mass of cylinder, m_1 = 65.1 g

Mass of liquid and cylinder combined, M = 120.5 g

Mass of liquid, m_2 = (M-m_1)=(120.5-65.1)g=55.4g

To calculate density of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Mass of liquid = 55.4 g

Volume of liquid = 49.3 mL = 49.3 cm^3    (Conversion factor:  1mL=1cm^3 )

Putting values in above equation, we get:

\text{Density of liquid}=\frac{55.4g}{49.3cm^3}\\\\\text{Density of liquid}=1.12g/cm^3

Hence, the density of liquid is 1.12g/cm^3

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3 years ago
An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .
grandymaker [24]

Answer: The specific heat of the unknown metal is 0.897J/g^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=5040 Joules

m= mass of substance = 86.8 g

c = specific heat capacity = ?

Initial temperature of the water = T_i

Final temperature of the water = T_f

Change in temperature ,\Delta T=T_f-T_i=(64.7)^0C

Putting in the values, we get:

5040=86.8\times c\times 64.7^0C

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4 0
3 years ago
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An electron can be added to halogen atom to force a halide ion with 8 valence electrons

<h3>What is an atom?</h3>

An atom can be defined as the smallest part of an element which can take part in a chemical reaction.

However whenever, an electron is added to halogen atom to force a halide ion with 8 different valence electrons

So therefore; an electron can be added to halogen atom to force a halide ion with 8 valence electrons

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2 years ago
Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of
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Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x

Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x

Total energy required = 2260x+x\times2.010\times(135-100)=2260x+70.35x=2330.35x

2330.35x=1048815\\x=450.068g

Hence, there was 450.068g of water in the pot.

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