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maria [59]
3 years ago
8

Chicago is about 2,800 km from Los Angeles. About how long would it take a jet plane flying 800 km/h to fly from Chicago to Los

Angeles? A. 2.0 h B. 2.5 h C. 3.0 h
Chemistry
1 answer:
BaLLatris [955]3 years ago
4 0

<u>Answer:</u> The time taken by jet plane to reach Los Angeles from Chicago is 3.5 hours

<u>Explanation:</u>

Speed is defined as the rate at which an object moves with respect to time.

To calculate the time taken for the given speed, we use the equation:

s=\frac{d}{t}

where,

s = speed of the jet plane = 800 km/hr

d = distance traveled = 2800 km

t = time taken by jet plane = ?

Putting values in above equation, we get:

800km/hr=\frac{2800km}{t}\\\\t=\frac{2800km}{800km/hr}=3.5hr

Hence, the time taken by jet plane to reach Los Angeles from Chicago is 3.5 hours

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What is restriction enzymes and what is it’s purpose?
Eddi Din [679]

Answer:  <u>Endonuclease enzymes used in molecular biology that cut DNA at specified points.</u>

Explanation:

Enzymes are specific protein types which bind to a substrate within a reaction, to increase the rate of reaction within the solution- they speed up the rate of reaction.

Restriction enzymes are bacteria-derived enzymes; these make cuts on deoxyribonucleic acid molecules or DNA. These are also called  restriction endonucleases. They are utilized in molecular biology for DNA cloning and sequencing and cut DNA into smaller pieces called fragments.

Restriction enzymes make directed cuts on DNA molecules. They precisely target sites on DNA to produce mostly identical or homogenous, discrete fragments of equal sizes, producing blunt or sticky ends. In order to do this, they recognize sequences of nucleotides that correspond with a complementary sequence on the endonuclease called restriction sites.

There are several kinds that may require cofactors (chemical or metallic compounds that aid in enzyme activity)  :

  • Type I: cleave far away from the recognition site; require ATP and  SAMe S-Adenosyl-L-Methionine

  • Type II: cleave near to the site; require Magnesium

  • Type III: cleave near to the site; require ATP which is not hydrolysed but  SAMe S-Adenosyl-L-Methionine is optional

  • Type IV: cleavage targeted to DNA that have undergone post transcriptional modification through certain types of methylation (addition of a methyl group)

7 0
3 years ago
The isotope Np-238 has a half life of 2.0 days if 96 grams of it were present on Monday how much will remain six days later
Kipish [7]

Answer:

12.02 g

Explanation:

From the question given above, the following data were obtained:

Half life (t½) = 2 days

Original amount (N₀) = 96 g

Time (t) = 6 days

Amount remaining (N) =..?

Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:

Half life (t½) = 2 days

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 2

K = 0.3465 /day

Therefore, the rate of disintegration of the isotope is 0.3465 /day.

Finally, we shall determine the amount of the isotope remaining after 6 days as follow:

Original amount (N₀) = 96 g

Time (t) = 6 days

Decay constant (K) = 0.3465 /day.

Amount remaining (N) =.?

Log (N₀/N) = kt / 2.303

Log (96/N) = (0.3465 × 6) / 2.303

Log (96/N) = 2.079/2.303

Log (96/N) = 0.9027

Take the anti log of 0.9027

96/N = anti log (0.9027)

96/N = 7.99

Cross multiply

96 = N × 7.99

Divide both side by 7.99

N = 96 /7.99

N = 12.02 g

Therefore, the amount of the isotope remaining after 6 days is 12.02 g

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3 years ago
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Answer: 2) 2HCl(sq) + CaCO3(s) CaCl2(sq) + CO2(g) + H2O (l) No of moles of CaCO3 = amount of the CaCO3 (g)/mw of CaCO3 (g/mole)= 0.8085 g/100 g/mole = 0.008085

Explanation:

7 0
2 years ago
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Vilka [71]

Answer:

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Explanation:

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3 0
3 years ago
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