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3 years ago

A mining crew extracted two different types of minerals from underground.

2 answers:

7 0

they could be different types of minerals, so one reacted to the energy differently than the other one. and pls answer my latest questions its due by 11:59 pm TONIGHT or i get kicked out of school.. pls help..

Oksana_A [137]3 years ago

4 0

Answer:

Mineral A's molecules had a high level of attraction before. As well as the molecules in mineral B. But, mineral A's molecules, probably had higher kinetic energy. Which means a higher temperature and faster-moving molecules. Which allowed it to spread more and since it spread more, it became a liquid. Mineral B might have had not that much kinetic energy so the temperature was colder and its molecules moved slower. That's why probably mineral A changed to a liquid while mineral B remained a Solid.

Explanation:

(Answer + Explanation are both the same thing btw)

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Answer:

A concentrated acid is an acid which is in either pure form or has a high concentration. Laboratory type sulfuric acid (about 98% by weight) is a concentrated (and strong) acid. A dilute acid is that in which the concentration of the water mixed in the acid is higher than the concentration of the acid itself.

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3 years ago

Read 2 more answers

Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f

pishuonlain [190]

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

What would happen to the gas pressure inside the cylinder if you do the following?

Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.

T₂ = 0.5 T₁. According to Gay-Lussac's law,

$\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}$

Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

$P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}$

P₂ in terms of the ideal gas equation is:

$P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}$

3 0

3 years ago