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kondor19780726 [428]
3 years ago
5

Given the unbalanced chemical equation below. What would be the coefficient of AgNO3?

Chemistry
1 answer:
OLga [1]3 years ago
8 0
2
I can’t really explain in words so I took a pic of the work I did (Ignore the worksheet and just look at what I wrote to balance the equation.)

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By reproduction which is the actions of mating, impregnating, and burthing of offspring.

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3 years ago
He equilibrium constant (KP) is 0.21 at a particular temperature for the reaction:
lorasvet [3.4K]

Explanation:

We will use the following equation,  

     Q_{p} = \frac{(P^{2}_{NO_{2}}{P_{N_{2}O_{4}}

(a) Now, we will calculate the value of Q_{p} as follows.

          Q_{p} = \frac{(0.225)^{2}}{0.134}

                      = 0.377

As, the K_{p} is 0.21 (which is smaller than the value) this means that equilibrium will shift to the left .

(b)   Q_{p} = \frac{(0.086)^{2}}{0.064}

                  = 0.1155

Here, Q_{p} is smaller than k_{p} which means that the equilibrium will shift to the right.

(c)   Q_{p} = \frac{(0.13)^{2}}{0.142}

                   = 0.119

Here, Q_{p} is smaller than k_{p} which means that the equilibrium will shift to the right .

(d)    Q_{p} = \frac{(0.173)^{2}}{0.142}

                  = 0.21  

Since, here the value of Q_{p} comes out to be equal to the given value. Therefore, this reaction is at equilibrium.

(e)  Q_{p} = \frac{(0.151)^{2}}{0.062}

                  = 0.36

Here, Q_{p} is higher than k_{p} the reaction will shift to the left.

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3 years ago
An electron-dot structure is a convenient method of representing
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Answer:

all electrons of the atom

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How many grams are there in 7.4 x 1023 molecules of H2SO4?
bazaltina [42]

7.4x10^23 = molecules of silver nitrate sample  

6.022x10^23 number of molecules per mole (Avogadro's number)  

Divide molecules of AgNO3 by # of molecules per mol  

7.4/6.022 = 1.229 mols AgNO3 (Sig Figs would put this at 1.3)  

(I leave off the x10^23 because they both will divide out)  

Use your periodic table to find the molar weight of silver nitrate.  

107.868(Ag) + 14(N) + 3(16[O]) = 169.868g/mol AgNO3  

Now multiply your moles of AgNO3 with your molar weight of AgNO3  

1.229mol x 169.868g/mol = 208.767g AgNO3

4 0
4 years ago
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