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Annette [7]
3 years ago
8

Help me in my this plzzz ​

Chemistry
2 answers:
eduard3 years ago
8 0

Answer:

a is oxidation

b is reduction

c is reduction

d is oxidation

hope it helps you

Talja [164]3 years ago
6 0
That the right question
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Chase buys a helium balloon for a friend's birthday. As he walks out of the store, it starts snowing.
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I think its A sry if im wrong

Explanation:

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3 years ago
What is the mass of the missing reactant in the following reactions, given that the reaction goes to completion? Do not include
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1) <span> 2.7 g water + 6,6 g carbon dioxide </span>→<span> 9.3 g carbonic acid.
According to </span><span>principle of mass conservation mass of reactants and products are the same after chemical reactio. 2,7 g + 6,6 g = 9,3 g.
2) </span><span>32.0 g sodium hydroxide + 16.0 g hydrofluoric acid --> 14,4 g water + 33.6 g sodium fluoride.
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Electrochemistry is defined as a science that deals with the relation of electricity to chemical changes and the interconversion
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Placing two metal strips in an electrolyte.
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2. Using the following data, calculate the average atomic mass of magnesium (give your answer to the nearest
Savatey [412]

Answer: 24.309amu

Explanation:

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3 years ago
In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) + H2O (g) ⇌
Oksana_A [137]

Answer: Equilibrium constant is 0.70.

Explanation:

Initial moles of  CO = 0.35 mole

Volume of container = 1 L

Initial concentration of CO=\frac{moles}{volume}=\frac{0.35moles}{1L}=0.35M

Initial moles of  H_2O = 0.40 mole

Volume of container = 1 L

Initial concentration of H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M

equilibrium concentration of CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M [/tex]

The given balanced equilibrium reaction is,

                            CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial conc.            0.35 M       0.40M       0     0

At eqm. conc.    (0.35-x) M   (0.40-x) M   (x) M    (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}

K_c=\frac{x\times x}{(0.40-x)(0.35-x)}

we are given : (0.35-x)= 0.18

x = 0.17

Now put all the given values in this expression, we get :

K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}

K_c=0.70

Thus the value of the equilibrium constant is 0.70.

5 0
3 years ago
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