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m_a_m_a [10]
3 years ago
8

When liquid heats up is it a chemical change

Chemistry
1 answer:
Mazyrski [523]3 years ago
5 0
Yes because of the gas combination.
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How do we turn petroleum into energy we use?
arlik [135]

You don't "turn" it into energy; petroleum HAS stored energy (chemical energy).However, you can turn it into ANOTHER TYPE OF ENERGY; usually this is done by burning the petroleum, and using it to drive machinery.

Since burning fuels is wasteful (the efficiency is limited, in theory, to the Carnot efficiency of a heat engine), other options are being explored, such as chemical reactions in a fuel cell. But such technology is not yet used on a large scale.

7 0
4 years ago
Phosphine (PH3) can be prepared by the reaction of calcium phosphide , Ca3P2: based on this equation : Ca3P2 + 6H2O —-> 3 Ca(
MrMuchimi

Taking into account the reaction stoichiometry, you can observe that:

  • one mole of Ca₃P₂ produces 2 mol of PH₃.
  • the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Ca₃P₂ + 6 H₂O  → 3 Ca(OH)₂ + 2 PH₃

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

  • Ca₃P₂:1 mole
  • H₂O: 6 moles
  • Ca(OH)₂: 3 moles
  • PH₃: 2 moles

The molar mass of the compounds is:

  • Ca₃P₂: 182 g/mole
  • H₂O: 18 g/mole
  • Ca(OH)₂: 74 g/mole
  • PH₃: 34 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Ca₃P₂: 1 mole ×182 g/mole= 182 grams
  • H₂O: 6 moles× 18 g/mole= 108 grams
  • Ca(OH)₂: 3 moles ×74 g/mole= 222 grams
  • PH₃: 2 moles ×34 g/mole= 68 grams

<h3>Correct statements</h3>

Then, by reaction stoichiometry, you can observe that:

  • one mole of Ca₃P₂ produces 2 mol of PH₃.
  • the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

4 0
2 years ago
What are the weighting laws?
lora16 [44]

Answer: A comprehensive legal term for uniform standards ascribed to the quantity, capacity, volume, or dimensions of anything. Legislation that adopts and mandates the use of a uniform system of weights and measures is a valid exercise of Police Power, and such laws are constitutional.

Explanation:

8 0
3 years ago
How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102
Ede4ka [16]

Answer: 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles

1 mole of hydrogen (H_2) = 2\times 6.023\times 10^{23}=12.05\times 10^{23} atoms

17.5 mole of hydrogen (H_2) = \frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25} atoms

There are 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0
3 years ago
Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the sys
8090 [49]

Answer:

Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.

Explanation:

The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:

Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}.

Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both [\mathrm{SO_2\, (g)}] and [\mathrm{O_2\, (g)}] will increase if the pressure is increased through compression. However, because \rm SO_2\, (g) and \rm O_2\, (g) have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient Q.

As a result, the increase in pressure will have no impact on the value of Q\!. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.

8 0
3 years ago
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