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alexgriva [62]
2 years ago
10

A constantly changing and developing resume means that you are always _______________ as a person and potential employee. A. cha

nging and developing B. failing C. succeeding D. remaining the same
Chemistry
1 answer:
abruzzese [7]2 years ago
4 0
I believe your answer is A “changing and developing”

Good luck!
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In the Energy and Specific Heat lab, how should the water bath be stabilized over the heat source? Select one:
Reika [66]

Answer:

c

Explanation:

5 0
3 years ago
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Why don't we discuss the results during the results section of the project?
snow_tiger [21]

Answer: The result section of the project contains your findings while carrying out your research or study.

Explanation:

The Results section of a research or study usually contains only the findings of your study or research.The findings which usually include

1. Data presented in tables, charts, graphs, and other figures.

2. A contextual analysis of this data explaining its meanings. Usually in sentences.

Our result gotten is not discussed in result section because every project or research work has a discussion page where every results or findings are discussed. The result section is expected to carry what you found.

4 0
3 years ago
What is a societal law
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Societal law is a rule which stretches out to ones in and by the public
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3 years ago
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VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu
SOVA2 [1]

Answer:

31.9 °C  

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

q = 12.35 cal

m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹  

ΔT  = 12.35/(2.406 °C⁻¹) = 5.13 °C

(b) Calculate T₂

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

 

6 0
3 years ago
a 160 milligram sample of a radioactive isotope decays to 10 kilograms in 12 years. what is the half life of this element
Shkiper50 [21]

Answer:

3 years

Explanation:

Given data:

Initial amount of sample = 160 Kg

Amount left after 12 years = 10 Kg

Half life = ?

Solution:

at time zero = 160 Kg

1st half life = 160/2 = 80 kg

2nd half life = 80/2 = 40 kg

3rd half life = 40 / 2 = 20 kg

4th half life = 20 / 2 = 10 kg

Half life:

HL = elapsed time / half life

12 years / 4 = 3 years

8 0
3 years ago
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