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postnew [5]
3 years ago
10

The decomposition of potassium chlorate yields oxygen gas. If the yield is 895%, how many grams of KClO3 are needed to produce 1

0.0 L of O2 ?
2KClO3(s) -- > 2KCl(s) + 3 O2(g)

Chemistry
1 answer:
harina [27]3 years ago
4 0

The answer is either 38 g or 40.g KCLO3. If it 95% instead of 895%.

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The energy released from 1 gram of uranium is more than 1 million times greater than the energy released from 3 grams of coal is True.

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In the given example, it is obvious that the energy released from 1 gram of uranium is more than that of the energy released from 3 grams of coal because the amount of energy released during nuclear fission is millions of times more efficient per mass than that of coal considering only \frac{1}{10}^{th} part of the original nuclei is converted to energy.

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How to prepare 1m of a solution​
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Explanation:

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2 years ago
Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid H3PO4 with
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Answer:

7.3 g (NH₄)₃PO₄

Explanation:

The balanced equation for the reaction is:

H₃PO₄ +  3 NH₃ ----> (NH₄)₃PO₄

To find the mass of ammonium phosphate ((NH₄)₃PO₄) produced, you need to (1) convert grams NH₃ to moles NH₃ (via the molar mass from the periodic table), then (2) convert moles NH₃ to moles (NH₄)₃PO₄ (via mole-to-mole ratio from balanced equation), and then (3) convert moles (NH₄)₃PO₄ to grams (NH₄)₃PO₄ (via molar mass from periodic table). Make sure to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 2 sig figs because the given value (2.5 grams) has 2 sig figs.

Molar Mass (NH₃): 14.01 g/mol + 3(1.008 g/mol)

Molar Mass (NH₃): 17.034 g/mol

Molar Mass ((NH₄)₃PO₄):

3(14.01 g/mol) + 12(1.008 g/mol) + 30.97 g/mol + 4(16.00 g/mol)

Molar Mass ((NH₄)₃PO₄): 149.096 g/mol

2.5 g NH₃          1 mole NH₃        1 mole (NH₄)₃PO₄             149.096 g
---------------  x  --------------------  x  ---------------------------  x  --------------------------
                            17.034 g             3 moles NH₃              1 mole (NH₄)₃PO₄

=  7.3 g (NH₄)₃PO₄

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2 years ago
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