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Whitepunk [10]
2 years ago
8

What indicates that two objects are in thermal equilibrium?

Chemistry
1 answer:
Leona [35]2 years ago
4 0

Answer:

B

Explanation:

equilibrium happens when two things are in complete balance and thermal is in reference to heat so two things are in thermal equilibrium once they have the same temperature.

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Arrange the following element, lithium, potassium, carbon and fluorine in increasing ionization energy
Juli2301 [7.4K]

Answer:

flourine(1681),carbon(1086),lithium(520) and potassium(419)

as we can see that the ionization energy of flourine is the highest than carbon than lithium and than potassium

Explanation:

i hope it will help you

4 0
2 years ago
Read 2 more answers
The average atomic mass of Eu is 151.96 amu. There are only two naturally occurring isotopes of europium, Eu with a mass of 151.
earnstyle [38]

Answer:

The percentage abundance of Eu isotopes are 52 %  and 48 % .

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope,:

% = x %

Mass = 151.0 amu

For second isotope :

% = 100  - x  

Mass = 153.0 amu

Given, Average Mass = 151.96 amu

Thus,

151.96=\frac{x}{100}\times {151.0}+\frac{100-x}{100}\times {153.0}

Solving for x, we get that:

x = 52 %

<u>Thus percentage abundance of Eu isotopes are 52 %  and 48 % .</u>

7 0
2 years ago
Robert has 4 grams of a substance and Jill has 10 grams of the same substance. Which of the following statements is true?
ELEN [110]
If Robert has 4 grams of a substance and Jill has 10 grams of the same substance <span>Jill's sample will weigh more than Robert's sample.</span>
5 0
2 years ago
Read 2 more answers
Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b.
Ilia_Sergeevich [38]

Answer:

a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe

b. Glu-Ala-Phe + Gly-Ala-Tyr

Explanation:

In this case, we have to remember which peptidic bonds can break each protease:

-) <u>Trypsin</u>

It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.

-) <u>Chymotrypsin</u>

It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.

With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the <u>"Lis"</u> and <u>"Arg"</u> (See figure 1).

In "peptide b", the peptidic bond that would be broken is the one in the <u>"Phe"</u> (See figure 2). The second amino acid that can be broken is <u>tyrosine</u>, but this amino acid is placed in the <u>C terminal spot</u>, therefore will not be involved in the <u>hydrolysis</u>.

3 0
3 years ago
Balance the following reaction:
kvv77 [185]
  • C_5H_8+13/2O_2—»5CO_2+4H_2O

Balanced one

  • 2C_5H_8+13O_2—»10CO_2+8H_2O

Moles of Pentyne

  • Given mass/Molarmass
  • 34/68
  • 0.5mol

Moles of H_2O

  • 8/2(0.5)
  • 4(0.5)
  • 2mol

1mol releases 241.8KJ

2mol releases 241.8(2)=483.6KJ

8 0
1 year ago
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