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RoseWind [281]
3 years ago
13

We can also use the equation for enthalpy change for physical phase changes. Consider the phase change H2O(l) → H2O(g). Calculat

e ΔHrxn.
Chemistry
2 answers:
BlackZzzverrR [31]3 years ago
4 0
<span>44.0 kJ is the answer</span>
Lostsunrise [7]3 years ago
3 0

The answer will be <u>44.0 kJ</u> .

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H E L P <br> P L E A S E<br> ;-;
aleksandr82 [10.1K]

Answer:

because it decompose

Explanation:

yeast is an organic compound

7 0
3 years ago
1. When this equation is balanced, Fe + 02 -.--&gt; Fe203, what is the
Vilka [71]

Fe + O2 → Fe2O3

After balancing the eq.

4Fe + 3O2 → 2Fe2O3

Hope this will help u mate :)

4 0
3 years ago
If I have 21 liters of gas held at a pressure of 78 atm and a temperature of 900 K, what will
shusha [124]

Answer:

30.33L

Explanation:

Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant  and Charles law states that the volume of a given mass of gas is directly proportional to the temperature provided the pressure remains constant

P1V1/T1 = P2V2/ T2

P1 =  78atm, V1 = 21L , T1 = 900K

P2 = 45atm, V2 = ? , T2 =750K

78× 21 / 900 = 45×V2 / 750

1638/900 = 45 V2 / 750

1638×750 = 900×45V2

1228500 = 40500V2

Divide both sides by 40500

1228500÷40500= V2

V2 = 30.33L

I hope this was helpful, please mark as brainliest

6 0
3 years ago
What type of mixture will not allow light to pass through
igor_vitrenko [27]
Suspension.  The particles are big enough for the eye to see, and will separate if left sitting.
5 0
3 years ago
How many moles of water can be produced with 4.3 moles of H2 and 5.6 moles of O2? Which reactant is limiting? How many moles of
xeze [42]

Answer:

Hydrogen H₂ will be the limiting reagent.

The excess reactant that will be left after the reaction is 3.45 moles.

4.3 moles of water can be produced.

Explanation:

The balanced reation is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • H₂: 2 moles
  • O₂: 1 mole
  • H₂O: 2 moles

To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?

moles of H_{2} =\frac{5.6 moles of O_{2} *2 mole of H_{2} }{1 mole of O_{2}}

moles of H₂= 11.2 moles

But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, <u><em>hydrogen H₂ will be the limiting reagent</em></u> and oxygen O₂ will be the excess reagent.

Then you can apply the following rules of three:

  • If by reaction stoichiometry 2 moles of H₂ react with 1 mole of O₂, 4.3 moles of H₂ will react with how many moles of O₂?

moles of O_{2} =\frac{1 mole of O_{2} *4.3 mole of H_{2} }{2 mole of O_{2}}

moles of O₂= 2.15 moles

The excess reactant that will be left after the reaction can be calculated as:

5.6 moles - 2.15 moles= 3.45 moles

<u><em>The excess reactant that will be left after the reaction is 3.45 moles.</em></u>

  • If by reaction stoichiometry 2 moles of H₂ produce 2 moles of H₂O, 4.3 moles of H₂ produce how many moles of H₂O?

moles of H_{2}O =\frac{2 moles of H_{2}O *4.3 mole of H_{2} }{2 mole of H_{2}}

moles of H₂O= 4.3 moles

<u><em>4.3 moles of water can be produced.</em></u>

8 0
3 years ago
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