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amm1812
3 years ago
9

For each of the following compounds, write the symbols of the ions in the compound, and the number of each ion in the molecule o

f that compound.
Cal2
Na2CO3
Ga(CIO3)3
CuF2
(NH4)3PO4
FeSO4
Mg(NO3)2
NH4NO2
KC2H3O2
Na2Cr2O7
Chemistry
1 answer:
Kay [80]3 years ago
8 0
Ca^2+ and I^-
Na+ and Co3^2-
Ga^3+ and ClO3
Cu^2+ and F-
NH4^- and PO4^3-
Fe2+ and (SO4)^2-
Mg2+ and NO3^-
NH4^+ and NO2^-
K^+ and (C2H3O2)^- {C2H3O2 is acetate}
Na^+ and Cr2O7^2-
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An atom has 9 electrons and 9 protons at the start . If it loses 2 electrons, the net charge on the atom will be ? If the atom i
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The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react
murzikaleks [220]

<u>Answer:</u> The value of K_{eq} is 4.84\times 10^{-5}

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of ammonia = \frac{0.0280}{1.00}=0.0280M

Concentration of oxygen gas = \frac{0.0120}{1.00}=0.0120M

The given chemical equation follows:

                  4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)

<u>Initial:</u>        0.0280        0.0120

<u>At eqllm:</u>    0.0280-4x   0.0120-3x   2x       6x

We are given:

Equilibrium concentration of nitrogen gas = 3.00\times 10^{-3}M=0.003

Evaluating the value of 'x', we get:

\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M

Now, equilibrium concentration of ammonia = 0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M

Equilibrium concentration of oxygen gas = 0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M

Equilibrium concentration of water = 6x=(6\times 0.0015)]=0.009M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}

Putting values in above expression, we get:

K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}

Hence, the value of K_{eq} is 4.84\times 10^{-5}

4 0
3 years ago
3. Burns from boiling water can be severe, caused by the transfer of energy from the boiling water to the
Free_Kalibri [48]

Answer:

Q = 3937.56  J

Explanation:

Heat transferred due to change in temperature is given by :

Q=mc\Delta T

c is the specific heat of water, c=4.18 J/g-°C

We have, m = 15 g, T_i=100^{\circ} C\ \text{and}\ T_f=37.2^{\circ} C

So,

Q=15\times 4.18\times (37.2-100)\\Q=-3937.56\ J

Hence, 3937.56  J of heat is transferred.

8 0
3 years ago
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