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Agata [3.3K]
3 years ago
10

Mg(s) + ½O2(g) → MgO(s) + 146 kcal/mole H2(g) + ½O2(g) → H2O(g), ΔH = -57.82 kcal/mole What type of reaction is represented by t

he previous two examples?
Endothermic
Exothermic
Chemistry
1 answer:
professor190 [17]3 years ago
4 0
Answer is: both reactions are exothermic.
<span>
In exothermic reactions, heat is released and enthalpy of reaction is less than zero (as it show second chemical reaction).
According to Le Chatelier's principle when the reaction is exothermic heat is included as a product (as it show first chemical reaction).</span>
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Which of the following is a chemical property of the noble gases?
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Noble gases are the elements located in the eighth group of the periodic table. These elements have the following properties:
They are odorless
They have single atoms (monoatomic)
They are colorless
They have very low chemical reactivity

The naturally occurring noble gases are helium, neon, argon, krypton, xenon and radon.
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Write the balanced chemical equation for the reaction
kykrilka [37]

Answer:

  1. 2C +O2 - CO

Explanation:

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6 0
2 years ago
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol
Kryger [21]

Answer:

0.44 moles

Explanation:

Given that :

A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.

The equilibrium constant K_c=  \dfrac{[CO][H_2]}{[H_2O]}

The equilibrium constant  K_c=  \dfrac{(0.17 )(0.17)}{0.74}

The equilibrium constant K_c=  0.03905

Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.

The equation for the reaction is :

H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \  \ \ \ \ \to0.17

Total mole of water now = 0.74+0.17

Total mole of water now = 0.91 moles

Again:

K_c=  \dfrac{[CO][H_2]}{[H_2O]}

0.03905 =  \dfrac{[0.17+x][x]}{[0.91 -x]}

0.03905(0.91 -x) = (0.17 +x)(x)

0.0355355 - 0.03905x = 0.17x + x²

0.0355355 +0.13095 x -x²

x² - 0.13095 x - 0.0355355 = 0

By using quadratic formula

x = 0.265  or   x = -0.134

Going by the value with the positive integer; x = 0.265 moles

Total moles of CO in the flask when the system returns to equilibrium is :

= 0.17 + x

= 0.17 + 0.265

= 0.435 moles

=0.44 moles (to two significant figures)

3 0
3 years ago
Can someone please help me on the second one? I have no idea what to do.
Aleks [24]

The reaction will produce 12.1 g Ag₂S.

<em>Balanced equation</em> = 2Ag + S ⟶ Ag₂S

<em>Mass of Ag₂S</em> = 10.5 g Ag × (1 mol Ag/107.87 g Ag) × (1 mol Ag₂S/2 mol Ag)

× (247.80 g Ag₂S/1 mol Ag₂S) = 12.1 g Ag₂S

8 0
3 years ago
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