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3 years ago

When wax is heated and changes to liquid how are the molecule affected

Chemistry

1 answer:

Alexus [3.1K]3 years ago

4 0

The molecules gain energy and the space between them decreases

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a sample of a substance has a density of 1.75 g/cm3 and a mass of 14.7 g. What volume does the sample occupy?

Zigmanuir [339]

The volume occupied by the sample is 8.4 cm³

<h3>What is density?</h3>

The density is the ratio of mass of the object to its volume.

Given is the mass m =14.7 g and the density is 1.75 g/cm³, then the volume will be

ρ = m/ V

V =m /ρ

Substitute the values, we get

V = 14.7 /1.75

V = 8.4 cm³

Thus, the volume of the sample is 8.4 cm³.

Learn more about density.

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3 years ago

2(NH4)₂Cr₂ What are the names of the atoms make up the above formula?

mars1129 [50]

Answer:

Nitrogen, Hydrogen, Chromium

Explanation:

N= Nitrogen

H= Hydrogen (it is not helium because helium is "He")

Cr= Chromium

6 0

2 years ago

Determinar los mililitros de solución 0.5 Normal de NaCl que se pueden preparar a partir de 3 L de solución de cloruro de sodio

MrMuchimi

Answer:

uhmmm english?

Explanation:

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3 years ago

Calculate the heat released when 454 grams of steam at 100 degrees C condenses.

sasho [114]

<h3>Answer:</h3>

1026.05 kJ

<h3>Explanation:</h3>

The Quantity of heat is calculated by multiplying the mass of a substance by specific heat capacity and the change in temperature.
However, condensation takes place at a constant temperature.
Therefore, Quantity of heat is calculated by multiplying mass by the latent heat of vaporization.

That is; Q = m × Lv

In this case we are given;

Mass of steam as 454 g

To calculate the amount of heat released we must know the value of latent heat of vaporization;

The latent heat of vaporization of steam is 2260 J/g

Therefore;

Heat released, Q = 454 g × 2260 J/g

=1026040 Joules

But, 1000 joules = 1 kilo-joules

Thus, Q = 1026.04 kJ

Hence, the amount of heat released is 1026.05 kJ

5 0

3 years ago

A 112 gram sample of an unknown metal was heated from 0.0C to 20.0C. The sample absorbed 1004 J of energy. What was the specific

worty [1.4K]

Answer:

The specific heat of the sample unknown metal is approximately 0.45 J/g °C.

General Formulas and Concepts:
<u>Thermodynamics</u>

Specific Heat Formula: $\displaystyle q = mc \triangle T$

<em>m</em> is mass (g)
<em>c</em> is specific heat capacity (J/g °C)
Δ<em>T</em> is the change in temperature

Explanation:

<u>Step 1: Define</u>

<em>Identify variables.</em>

<em>m</em> = 112 g

Δ<em>T</em> = 20.0 °C

<em>q</em> = 1004 J

<u>Step 2: Solve for </u><u><em>c</em></u>

Substitute in variables [Specific Heat Formula]: $\displaystyle 1004 \ \text{J} = (112 \ \text{g})(c)(20.0 \ ^{\circ} \text{C})$
Simplify: $\displaystyle 1004 \ \text{J} = (2240 \ \text{g} \ ^\circ \text{C})c$
Isolate <em>c</em>: $\displaystyle c = 0.448214 \ \text{J} / \text{g} \ ^\circ \text{C}$
Round [Sig Figs]: $\displaystyle c \approx 0.45 \ \text{J} / \text{g} \ ^\circ \text{C}$

∴ specific heat capacity <em>c</em> is equal to around 0.45 J/g °C.

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Topic: AP Chemistry

Unit: Thermodynamics

5 0

3 years ago