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dolphi86 [110]
3 years ago
14

During an experiment,what is the factor you are testing called

Chemistry
1 answer:
Step2247 [10]3 years ago
4 0

Answer:

the answer is variable

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Ammonium perchlorate NH4ClO4 is a powerful solid rocket fuel, used in the Space Shuttle boosters. It decomposes into nitrogen N2
Bas_tet [7]

Answer:

1.3 mol H₂O

Explanation:

Let's consider the decomposition reaction of ammonium perchlorate.

NH₄ClO₄(s) → 1/2 N₂(g) + 1/2 Cl₂(g) + O₂(g) + 2 H₂O(g)

As we can see in the balanced equation, the molar ratio of ammonium perchlorate to water is 1:2. The moles of water produced by the reaction of 2.5 mol of ammonium perchlorate.

2.5 mol NH₄ClO₄ × (2 mol H₂O / 1 mol NH₄ClO₄) = 1.3 mol H₂O

3 0
3 years ago
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Magnesium oxide decomposes into magnesium and oxygen. if 12.09 g of magnesium oxide decomposes to form 7.29 g of magnesium, what
Vlada [557]
 the mass of oxy gas would be 4.03 did i calculate that right?
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4 years ago
Help! I don't have the things to do this and its due TODAY! Could someone help me?
vodomira [7]

Answer:

--- The hot air inside the bottle creates an expansion and when the hot water is applied a contraction occurs, thus causing the balloon to be sucked.

--- An internal pressure is produced, joining the heat begins to decrease. As a consequence, the air molecule causes the balloon to enter the bottle thanks to the effect of the external pressure.

Explanation:

I hope I have helped with something.

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7 0
4 years ago
Determine the number of moles in 3.13 × 10²³ formula units of MgCl₂.
love history [14]

Answer:

<h2>0.52 moles </h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{3.13 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{3.13}{6.02}  \\  = 0.519933

We have the final answer as

<h3>0.52 moles</h3>

Hope this helps you

3 0
3 years ago
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