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leonid [27]
3 years ago
5

State the Constituents of the following alloys:magnalium and bronze​

Chemistry
1 answer:
____ [38]3 years ago
8 0
Magnalium: Magnesium and Aluminum

Bronze: Copper, Tin, Arsenic, Phosphorus, Aluminum, Manganese and Silicon (whichever you learned in class from those)
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The balanced chemical equation for the reaction of copper (Cu) and silver nitrate (AgNO3) is shown below.
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Remark
The balance numbers in front of Ag and AgNO3 are both 2. That number is in moles.

Rule: if the moles are the same in the equation, then whatever you are given for one, will be the same for the other. So you have 0.854 moles of Ag. You will also have 0.854 moles of AgNO3 

Answer: 0.854 <<<<<
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Answer:

B. Cell Type

Explanation:

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Which equation is balanced?
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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

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

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

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