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Komok [63]
2 years ago
11

How many square centimeters are in an area of 8.50 inch^2

Chemistry
1 answer:
Paha777 [63]2 years ago
5 0

Explanation:

Given parameters:

  Dimension = 8.5in²

we are to convert from in² to cm²

 in² = in x in

 cm² = cm x cm

   1 in = 2.54cm

Using dimensional analysis;

   8.5in² x \frac{2.54cm}{1in}  x \frac{2.54cm}{1in}

        =  54.84cm²

We solved this problem using dimensional analysis where we multiplied the unit with 1.

learn more:

Conversion brainly.com/question/555814

#learnwithBrainly

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The picture below shows a specific nuclear reaction occurring. What best describes the unlabeled particle shown at the bottom of
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Maybe it A but im not sure
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Which type of fuel can only be used in certain geographic area? A. natural gas B. nuclear C. solar D. Biomass
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A

Explanation:

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What is the formula used to calculate the heat required to warm or cool one phase of matter
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Answer:

SPECIFIC HEAT CAPACITY

Explanation:

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The compound ncl3 is nitrogen trichloride, but alcl3 is simply aluminum chloride. why
AveGali [126]
The answer to this question would be:
NCl3 is a molecular compound (two or more nonmetals), and therefore in its name prefixes indicate the number of each type of atom. so NCl3 is nitrogen trichloride<span>.
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5 0
3 years ago
Read 2 more answers
Calculate the heat absorbed by the water in a calorimeter when 175 grams of lead cools from 125.0°C to 22.0°C. The specific heat
Iteru [2.4K]

Answer:

Q = 233.42 J

Explanation:

Given data:

Mass of lead = 175 g

Initial temperature = 125.0°C

Final temperature = 22.0°C

Specific heat capacity of lead = 0.01295 J/g.°C

Heat absorbed by water = ?

Solution:

Heat  absorbed by water is actually the heat lost by the metal.

Thus, we will calculate the heat lost by metal.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 22.0°C - 125.0°C

ΔT = -103°C

Q = 175 g × 0.01295 J/g.°C×-103°C

Q = -233.42 J

Heat absorbed by the water is 233.42 J.

5 0
3 years ago
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