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UNO [17]
2 years ago
11

Anh measured the temperature of a pond near his house. Before he left for school, the water in the pond was 18 degrees celsius.

When he came home from school, the temperature of the pond was higher than it was in the morning. What happened to the water molecules while he was at school?
A) The molecules get closer together.
B) the molecules started moving faster.
C) The water lost its molecules.
D) The molecules became larger.
Chemistry
1 answer:
Hitman42 [59]2 years ago
3 0

As the temperature increases, B) the molecules started moving faster.

Explanation:

The temperature of a substance is a measure of the average kinetic energy of the particles in a substance. In particular, it can be found that the temperature is directly proportional to the average kinetic energy of the particles:

T\propto KE

The kinetic energy of a particle is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its speed

This means that the higher the temperature of a substance, the greater the speed of the particles in the substance.

Therefore, if we apply this concept to this problem, we infer that as the temperature of the water in the pond gets higher, the speed of the molecules inside the water increases, which means that the molecules are moving faster.

Therefore, the correct answer is

B) the molecules started moving faster.

Learn more about temperature:

brainly.com/question/1603430

brainly.com/question/4370740

#LearnwithBrainly

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How could installing new technology, such as scrubber machines, affect the factories required to install them? Name a positive a
Sindrei [870]

Answer:

Installing new technology, such as scrubbers, in factories will decrease their harmful emissions. This helps improve the safety of the surrounding community and the workers. But this technology is expensive and requires time and effort to install.

Explanation:

Hope this helped :)

5 0
2 years ago
How many acidic protons are there in 0.6137 g of KHP?
andreyandreev [35.5K]

0.6137 g of KHP contains 1.086 × 10^21 acidic protons.

Number of moles of KHP = mass of KHP/molar mass of KHP

Molar mass of KHP = 204.22 g/mol

Mass of KHP = 0.6137 g

Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP

Now,  1 each molecule of KHP contains 1 acidic proton.

For 0.003 moles of KHP there are; 0.003 × 1 × NA

Where NA is Avogadro's number.

So;  0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23

= 1.086 × 10^21 acidic protons.

Learn more: brainly.com/question/16672114

7 0
2 years ago
The bonds in the compound MgSO4 can be described as
Butoxors [25]
C. Sulfur and oxygen (non metals) forms a covalent bond while the magnesium (a metal) will react with both non metals to form an ionic bond
7 0
2 years ago
The density of two liquids, A and B, are 1000. kg/m3 and 600. kg/m3, respectively. The two liquids are mixed in a certain propor
Varvara68 [4.7K]

Answer:

Mass of liquid B = 271.2 gram

Explanation:

Given:

Density of liquid A = 1000 kg/m³

Density of liquid B = 600 kg/m³

Density of mixture = 850 kg/m³

Mass of mixture = 1 kg

Assume:

Volume of liquid A = Va

Volume of liquid B = Vb

So,

Volume of mixture = Va + Vb

Mass of liquid A = 1000(Va)

Mass of liquid B = 600(Vb)

Mass of mixture = Mass of liquid A + Mass of liquid B

1 = 1000(Va) + 600(Vb)

Volume of mixture = 1 / 850

So,

(1/850) = Va + Vb

Vb = (1/850) - Va

1 = 1000(Va) + 600[(1/850) - Va]

Va = 7.25 × 10⁻⁴

Vb = (1/850) - Va

Vb = (1/850) - [7.25 × 10⁻⁴]

Vb = 4.25 × 10⁻⁴

Mass of liquid B = 600(Vb)

Mass of liquid B = 600(4.25 × 10⁻⁴)

Mass of liquid B = 271.2 gram

4 0
2 years ago
Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC
Salsk061 [2.6K]

<u>Answer:</u> The \Delta G for the reaction is 54.6 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

  • To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

  • The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm

Putting values in above equation, we get:

K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92

  • To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 22.92

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol

Hence, the \Delta G for the reaction is 54.6 kJ/mol

7 0
2 years ago
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