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KengaRu [80]
3 years ago
9

What changes within the star A lead to the red giant B?

Chemistry
1 answer:
Lady_Fox [76]3 years ago
8 0

Answer: The fusion of hydrogen to form helium

Explanation:

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A 200-gram sample of a radioactive nuclide is left to decay for 60 hours. At the end of 60 hours, only 25 grams of the sample is
Goryan [66]
Half life is the time taken by a radioactive isotope to decay by half its original mass. 
The original mass is 200 g
Time taken is 60 hours
Final mass is 25 g
Therefore;
Final mass = Original mass × (1/2)^n; where n is the number of half lives.
25 = 200 (1/2)^n
1/8 = (1/2)^n
    n = 3 
Three half lives = 60 hours
1 half lives = 20 hours
Therefore; the half life of the radioactive nucleus is 20 hours
5 0
3 years ago
How many atoms of hydrogen combine with one carbon atom to form methane
bonufazy [111]

Answer:four hydrogen atoms

Explanation:

4 0
3 years ago
Sevin, the commercial name for an insecticide used to protect crops such as cotton, vegetables, and fruit, is made from carbamic
xeze [42]

Answer:

Explanation:

Ratio of mass of C , N , H and O

= .8007 :0.9333:0.2016:2.133

Ratio of moles of C , N , H and O

= .8007/12 : .9333 / 14 : 0.2016 / 1 : 2.133/16

= .0667 : .0667: .2016 : .1333

= .0667 / .0667 : .0667 / .0667 : .2016 /.0667 : .1333 / .0667

= 1 : 1 : 3: 2

Hence empirical formula = CNH₃O₂

7 .

Weight of titanium Ti = 1.916 g

Weight of oxygen = 3.196 - 1.916 = 1.28 g

Ratio of weight of Ti and O

= 1.916 : 1.28

Ratio of moles  of Ti and O

1.916/48 : 1.28/16  [ Molecular  weight of Titanium is 48 ]

= .04 : .08

= .04/.04 : .08/.04

= 1 :2 .

Empirical formula

TiO₂

5 0
2 years ago
What two elements are present in the compounds in the last two rows of table 4​
castortr0y [4]

Answer:could u expand on the question

Explanation:

8 0
3 years ago
An electrochemical cell is powered by the half reactions shown below.
andrezito [222]
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation. 

To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign. 

Pb(s) --> Pb2+ +2e-      E0 = +0.13 V
Ag+ + e-  ---> Ag           E0 = +0.80 V

Adding up the E0's would yield an overall electric cell potential of +0.93 V.
7 0
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