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lorasvet [3.4K]
3 years ago
10

What happens for a star to enter old age?​

Chemistry
2 answers:
yaroslaw [1]3 years ago
6 0

Answer: for a star to start the process of become an old star it has to burn all the hydrogen in the star.

Explanation:

I hope this helps if not sorry :(

Have a good day :)

Stay motivated

Talja [164]3 years ago
4 0

Answer:

Because the interstellar medium is 97% hydrogen and 3% helium, with trace amounts of dust, etc., a star primarily burns hydrogen during its lifetime. A medium-size star will live in the hydrogen phase, called the main sequence phase, for about 50 million years. Once hydrogen fuel is gone, the star has entered “old age.”

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A chemical equation frequently indicates all of the following except the...
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Answer:

i think the answer is D.

Explanation:

for the chemical equation in all the other answers are true so therefor d is not.

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What is the total number of moles of the reactants in the following reaction?
exis [7]
Assuming that 2H2O(l) is the product, the reactants would be 2H2(g) and O2(g). The number of moles in the reactants would be 2 moles of H2 and 1 mole of O2.
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The temperature of a 500 ml sample of gas increases from 150 k to 300 k. what is the final volume of the sample of gas, if the p
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5 0
3 years ago
How many chlorine atoms would be in 6.02 X 10^23 units of gold III chloride
Pepsi [2]

Answer:

The number of chlorine atoms present in 6.02 \times 10^{23} units of gold III chloride is 18.066 \times 10^{23}

Explanation:

Formula of Gold (III) chloride: AuCl_{3}

<em>Avogadro Number</em> : Number of particles present in <u>one mole</u><u> </u>of a substance.

{N_{0}} =6.022 \times 10^{23}

Using,

n(moles)=\frac{Given\ number\ of\ particles}{N_{0}}

n =\frac{6.02\times 10^{23}}{6.022\times 10^{23}}

= 1 mole(0.9999 , nearly equal to 1 )

The given Gold III chloride sample is 1 mole in amount.

6.022 \times 10^{23}  = 1 mole of AuCl_{3}

In this Sample,

1 mole of AuCl_{3} will give = 3 mole of Chlorine atoms

1 mole of Cl contain = 6.022 \times 10^{23}

3 mole of Cl contain = 6.022 \times 10^{23}\times 3

3 mole of Cl contain =18.066 \times 10^{23}

So,

The number of chlorine atoms present in 6.02 \times 10^{23} units of gold III chloride is 18.066 \times 10^{23}

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Sodium hydrogen carbonate and citric acid
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Mixing all these would create sodium hydrogen carbonate (baking soda)
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