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Ahat [919]
3 years ago
5

Which equation is used to calculate the specific heat of the metal?

Chemistry
1 answer:
algol [13]3 years ago
3 0

i dont know gjchhhhhhhhhhhhhhhhhhhhhhhhh

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Write balanced reactions for the complete combustion of hydrogen. Express your answer as a chemical equation. Identify all of th
Fantom [35]

Answer:

Explanation:

A combustion  involves the reaction of a fuel with oxygen (O₂). During the reaction of combustion of hydrogen (H₂), H₂ reacts with O₂ to form water (H₂O). The <em>balanced chemical equation</em> is the following:

2 H₂(g) + O₂(g) → 2 H₂O(g)

According to the chemical equation, 2 moles of H₂O are obtained from the reaction of 2 moles of H₂ with 1 mol of O₂. All reactants and products are in the gaseous phase.

7 0
3 years ago
2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

8 0
3 years ago
Vocabulary crossword puzzle properties of minerals
goblinko [34]

Answer: what were

Explanation:

4 0
2 years ago
Read 2 more answers
How is burning magnesium different than burning methane
olya-2409 [2.1K]

You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.

Explanation:

To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:

2 Mg (s) + O₂ (g) → 2 MgO + heat

CH₄ (g) + 2  O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat

However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).

Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)

2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)

So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.

we have used the following notations:

(s) - solid

(g) - gas

(l) - liquid

Learn more about:

combustion reactions

brainly.com/question/13824679

#learnwithBrainly

6 0
3 years ago
When a mercury-202 nucleus is bombarded with a neutron, a proton is ejected. What element is formed?
Galina-37 [17]

That will make a gold-202 nucleus.

<h3>Explanation</h3>

Refer to a periodic table. The atomic number of mercury Hg is 80.

Step One: Bombard the \displaystyle ^{202}_{\phantom{2}80}\text{Hg} with a neutron ^{1}_{0}n. The neutron will add 1 to the mass number 202 of ^{202}_{\phantom{2}80}\text{Hg}. However, the atomic number will stay the same.

  • New mass number: 202 + 1 = 203.
  • Atomic number is still 80.

^{202}_{\phantom{2}80}\text{Hg} + ^{1}_{0}n \to ^{203}_{\phantom{2}80}\text{Hg}.

Double check the equation:

  • Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
  • Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.

Step Two: The ^{203}_{\phantom{2}80}\text{Hg} nucleus loses a proton ^{1}_{1}p. Both the mass number 203 and the atomic number will decrease by 1.

  • New mass number: 203 - 1 = 202.
  • New atomic number: 80 - 1 = 79.

Refer to a periodic table. What's the element with atomic number 79? Gold Au.

^{203}_{\phantom{2}80}\text{Hg} \to ^{202}_{\phantom{2}79}\text{Au} + ^{1}_{1}p.

Double check the equation:

  • Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
  • Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.

A gold-202 nucleus is formed.

6 0
3 years ago
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