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2 years ago

The law of physics which states that energy can't be created or destroyed but can transfer or change forms is known as

Chemistry

2 answers:

Nat2105 [25]2 years ago

8 0

The law of conservation energy..

Sedaia [141]2 years ago

5 0

The Law of Conservation of Energy

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Which method is suitable for collection a a<br>carbon dioxide.Explain

mel-nik [20]

Answer:

Carbon dioxide can be collected over water. Carbon dioxide is slightly soluble in water and denser than air, so another way to collect it is in a dry, upright gas jar.

Explanation:

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2 years ago

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Elanso [62]

Answer:

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2 years ago

HELP IM AB TO CRY

Lemur [1.5K]

Answer:

1.23 × 10³ N

Explanation:

Step 1: Given and required data

Mass of the person (m): 125 kg

Acceleration due to the gravitational force (g): 9.81 m/s²

Step 2: Calculate the force acting between the Earth and a 125-kg person standing on the surface of the Earth

We will use Newton's second law of motion.

F = m × g

F = 125 kg × 9.81 m/s²

F = 1.23 × 10³ N

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2 years ago

Some metals corrode very readily. Metals from which groups might be used to provide a protective coating? Justify the answer

Andreyy89

Explanation:

Cadmium, nickel, chromium, and silver are sometimes used as protective platings. Metals have a wide range of corrosion resistance. The most active metals (i.e., those that tend to lose electrons easily)--such as magnesium and aluminum--corrode easily and are listed at the top of Table 2-1.

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3 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$