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navik [9.2K]
3 years ago
8

3.80 moles of oxygen are used up in the reaction. How many moles of water are produced?5.How many grams of oxygen does it take t

o produce 9.60 moles of carbon dioxide?6.How many moles of carbon dioxide are produced with using 5.80 grams of acetylene?7.How many grams of oxygen are required to burn 13.50 grams of acetylene?
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
7 0

Answer:

2 KClO3 (s) = 2 KCl (s) + 3 O2 (g)

2.5 g x g

Explanation:

x g O2 = 2.5 g KClO3 x (1 mol KClO3) x (3 mol O2) x (32 g O2) = 0.98 g O2

(122.5 g KClO3) (2 mol KClO3) (1 mol O2)

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

2.5 g x g

x g KCl = 2.5 g KClO3 x (1 mol KClO3) x (2 mol KClO3) x (74.5 g KCl) = 1.52 g KCl

(122.5 g KClO3) (2 mol KClO3) (1 mol KCl)

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

x mol 10 mol

x mol KClO3 = 10 mol O2 x (2 mol KClO3) = 6.7 mol KClO3

(3 mol O2)

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have a good day :)

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2 years ago
What was the main reason why Alfred Wegener's theory about the movement of the continents was dismissed in 1912?
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His map of the ocean floor did not support his theory

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3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
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The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

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3 years ago
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