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jek_recluse [69]
3 years ago
6

This happens in massive star which convert hydrogen into helium? ​

Chemistry
1 answer:
katovenus [111]3 years ago
6 0

Answer:

nucleosynthesis. A star's mass determines what other type of nucleosynthesis occurs in its core (or during explosive changes in its life cycle).

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What are the formulas of the acids: <br>AsO4<br>CIO4 ( l )<br>S ( ll )<br>F ( l )<br>PO4 (lll)​
SVETLANKA909090 [29]
Mg3(AsO4)2
Ca(ClO4)2
[S (II) not sure]
[F (I) not sure]
PO₄³

Sorry I don’t know all of them, good luck though! :)
4 0
3 years ago
Which of the following is an example of potential energy?
maxonik [38]

Answer: the person standing up

Explanation:

the person has the potential to fall or move from position

6 0
3 years ago
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Complete the mechanism for the reaction of 2-butanol in sulfuric acid at 140 °C by adding any missing atoms, bonds, charges, non
Svetlanka [38]

Answer:

2-Butene

Explanation:

The first step is the <u>ionization</u> of the acid to produce the hydronium ion. Then the OH will attack this ion to produce a <u>charged species</u> that can be stabilized when <u>H2O is produced</u>.

Then an <u>elimination</u> takes place to produce the more <u>substituted alkene</u> 2-butene and the <u>hydronium ion</u> is gain produced.

4 0
3 years ago
How many moles are present in 102.55g of water (H20)
natulia [17]
About 5 moles of water are present. To be precise it’s 5.666666667 moles of H2O
6 0
3 years ago
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Find the pH of a 0.010 M HNO2 solution.
lidiya [134]
Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol

M (molarity) = 0.010 M (Mol/L)


Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = 5.0*10^{-4}
\alpha^2 (degree\:of\:ionization) = ?

Ka = M * \alpha^2
5.0*10^{-4} = 0.010* \alpha^2
0.010\alpha^2 = 5.0*10^{-4}
\alpha^2 = \frac{5.0*10^{-4}}{0.010}
\alpha^2\approx500*10^{-4}

\alpha\approx\sqrt{500*10^{-4}}
\alpha \approx 2.23*10^{-3}

Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:

[ H_{3} O^+] = M* \alpha
[ H_{3} O^+] = 0.010* 2.23*10^{-3}
[ H_{3} O^+] \approx 0.0223*10^{-3}
[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:
log10(2.23) ≈ 0.34
pH = ?
[ H_{3} O^+] = 2.23*10^{-5}

Formula:
pH = - log[H_{3} O^+]

Solving:
pH = - log[H_{3} O^+]
pH = -log2.23*10^{-5}
pH = 5 - log2.23
pH = 5 - 0.34
\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark

Note:. The pH <7, then we have an acidic solution.
6 0
3 years ago
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