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Pavlova-9 [17]
3 years ago
11

If you are given time and distance, you can determine power if you know

Chemistry
1 answer:
qwelly [4]3 years ago
7 0

Answer:

Yeah force

Yeah Energy

Yeah Watts

Yeah Joules

Explanation:

1. (Force * distance) /time = Power

2. Energy/time = Power

3. Cause Watts = Power

4. Joules is just the unit of energy

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How much heat is needed to melt 100.0 grams of ice that is already at 0°C?
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A. The heat is needed to melt 100.0 grams of ice that is already at 0°C is +33,400 J.

<h3>What is Specific heat capacity?</h3>

Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass.

<h3>Heat needed to melt the cube of ice</h3>

The heat is needed to melt 100.0 grams of ice that is already at 0°C is calculated as follows;

Q = mL

where;

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Q = 100 x 334

Q = 33,400 J

Thus, the heat is needed to melt 100.0 grams of ice that is already at 0°C is +33,400 J.

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6 0
11 months ago
In the gravimetric method, is it important that you dissolve your sample in exactly 10 ml of water? why or why not?.
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When utilizing the gravimetric method, it is crucial to completely dissolve your sample in 10 mL of water. A quantitative technique called gravimetric analysis employs the selective precipitation of the component under study from an aqueous solution.

A group of techniques known as gravimetric analysis are employed in analytical chemistry to quantify an analyte based on its mass. Gravimetric analysis is a quantitative chemical analysis technique that transforms the desired ingredient into a substance (of known composition) that can be extracted from the sample and weighed. This is a crucial point to remember.

Gravimetric water content (g) is therefore defined as the mass of water per mass of dry soil. To calculate it, weigh a sample of wet soil, dry it to remove the water, and then weigh the dried soil (mdry). Dimensions of the sample Water is commonly forgotten despite having a density close to one.

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8 0
1 year ago
ODIO POD
miss Akunina [59]

Answer:

\boxed{\text{40 mol Al}}

Explanation:

            Al₂O₃ ⟶ 2Al + 3O₂

n/mol:     20

\text{Moles of Al} = \text{20 mol Al$_{2}$O$_{3}$}\times \dfrac{\text{2 mol Al}}{\text{1 mol Al$_{2}$O$_{3}$}}= \textbf{40 mol Al}\\\text{You can produce }\boxed{\textbf{40 mol Al}}

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