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lesantik [10]
3 years ago
14

1 point

Chemistry
1 answer:
Allushta [10]3 years ago
5 0

Answer:

<h2>1 atm</h2>

Explanation:

The final pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

From the question we have

P_2 =  \frac{10 \times 2}{20}  =  \frac{20}{20}  \\

We have the final answer as

<h3>1 atm</h3>

Hope this helps you

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What is the atom inventory for the following equation after it is properly balanced? ____NaOH + ____CuCl2 Imported Asset ____NaC
faltersainse [42]

Answer:

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2;

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

Explanation:

NaOH + CuCL2 —> NaCl + Cu(OH)2

The balanced equation can be achieved by doing the following:

There are 2 oxygen and 2 hydrogen atom on the right side. This is balanced by putting 2 in front of NaOH as shown below:

2NaOH + CuCL2 —> NaCl + Cu(OH)2

This makes Na to be unbalanced. Now to balance Na, put 2 in front of NaCl as illustrated below

2NaOH + CuCL2 —> 2NaCl + Cu(OH)2

Now the equation is balanced.

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

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3 years ago
Please help me with this
Zina [86]

Answer:

C

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3 years ago
Read 2 more answers
HELP ASAP!!! 2C4H10(g) + 1302 ===&gt; 8CO2(g) + 10H2O(g)
Ivan

Answer:

30.34g (corrected to 4 significant figures).

Explanation:

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Since O2 is in excess and butane is the limiting reagent, the no. of moles of carbon dioxide produced depends on the no. of moles of butane reacted.

From the equation, the mole ratio of butane:Carbon dioxide = 2: 8 = 1: 4,

meaning 1 mole of butane gives 4 moles of CO2.

Using this ratio,

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mass of CO2 produced =   0.689655 x (12.0+16.0x2)

=30.34g (corrected to 4 significant figures).

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Answer:

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Fear factor is nonsense because fear does not exist in plants but they still compete for sunlight and water.

Factory factor is nonsense as well because this term does not exist in biology.

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I am not positive this is what you want but here lol:

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