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Naddik [55]
3 years ago
10

How does repeatability have an effect on scientific research?

Chemistry
1 answer:
Bas_tet [7]3 years ago
5 0
In science it is best to continue research, and rinse and repeat. This allows for a stronger hypothesis if you're results are the same every time, or change your hypothesis if you stumble upon new results. <span />
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Classify each of the following as energy primarily transferred as HEAT or energy primarily transferred as WORK.
IgorLugansk [536]
number 1 is work number 2 is heat
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How can scientists use stoichiometry to ensure that commercial products contain the components that manufacturers claim they do?
Rina8888 [55]

Answer:

I know you have been waiting awhile for this question to be answered :)

Stoichiometry is used in industry quite often to determine the amount of materials required to produce the desired amount of products in a given useful equation. Each one of these products requires stoichiometry. There would be no products from these industries without chemical stoichiometry.

Explanation:

Hopefully this helps :D

Sorry you had to wait so long :(

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Is oxygen gas (O2) a compound or an element? Is hydrogen gas (H2) a compound or an element?
Alinara [238K]
They're both elements
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Compare and contrast how wind glaciers abrade rock?
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A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
Lostsunrise [7]

Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

                     mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

5 0
3 years ago
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