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Svetach [21]
3 years ago
5

How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of alu

minum is 0.90 J/g° C?
Chemistry
1 answer:
Sergio039 [100]3 years ago
5 0

Answer:

297

Explanation:

55-22=33

Then use 33 *10*0.9

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Do the gas laws apply to liquids ?
Contact [7]

Answer:

The Ideal Gas Law cannot be applied to liquids. The Ideal Gas Law is #PV = nRT#. That implies that #V# is a variable. But we know that a liquid has a constant volume, so the Ideal <u><em>Gas Law cannot apply to a liquid.</em></u>

Explanation:

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Atoms are the smallest particles of a compound that still retain the properties of that
N76 [4]

Answer:

the answer is true

Explanation:

it is the smallest particle in an element that takes part in a chemical reaction

5 0
2 years ago
Enter a balanced equation for the reaction between solid nickel(II)(II) oxide and carbon monoxide gas that produces solid nickel
velikii [3]

Answer: A balanced equation for the given reaction is NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g).

Explanation:

The reaction equation will be as follows.

NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g)

Number of atoms on the reactant side is as follows.

  • O = 2
  • C = 1

Number of atoms on the product side is as follows.

  • Ni = 1
  • O = 2
  • C = 1

Since number of atoms on both the reactant and product sides are equal. Hence, the reaction equation is balanced.

Thus, we can conclude that a balanced equation for the given reaction is NiO(s) + CO \rightarrow Ni(s) + CO_{2}(g).

7 0
3 years ago
11. The oxyfuel flame was used for fusion welding as early as the first half of the
Firlakuza [10]

The Oxyfuel gas or flame refers to a group of welding processes that use the flame produced by the combination of a fuel gas and oxygen as the source of heat.

<u>Explanation:</u>

  • Oxy-fuel welding is a process that utilizes fuel gases and oxygen to weld metals. Oxyfuel gas or flame refers to a group of welding processes that utilize the flame delivered by the blending of fuel gas and oxygen as the source of heat.
  • This flame is utilized for cutting and welding of two metallic pieces. This is done due to the heat produced by cutting and welding of two metallic pieces together by heating to the melting point.
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7 0
3 years ago
Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat
Mila [183]

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values ​​of the atomic mass of each element that form the compounds:

  • N: 14 g/mol
  • H: 1 g/mol
  • O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

  • NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
  • O₂: 2*16 g/mol= 32 g/mol
  • NO: 14 g/mol + 16 g/mol= 30 g/mol
  • H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

  • NH₃: 4 moles
  • O₂: 5 moles
  • NO: 4 moles
  • H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

  • NH₃: 4 moles*17 g/mol= 68 g
  • O₂: 5 moles*32 g/mol= 160 g
  • NO: 4 moles*30 g/mol= 120 g
  • H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }

mass of O₂≅17.65 g

<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>

3 0
3 years ago
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