3 years ago

1. Predict the products for the given double replacement reaction: CaCl2 + AgNO3 -->

Chemistry

1 answer:

zheka24 [161]3 years ago

4 0

Answer:

1. Ca(NO3)2 and AgCl3

2. AgCl3

3. Ca(NO3)2

4. a solid that forms out of a solution

Explanation:

3CaCl2 + 2Ag(NO3)3 --> 3Ca(NO3)2 + 2AgCl3

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What is the mass of 40.34 x 10^23 H2 molecules at stp?

aev [14]

Every mol=6.22*10^23 <span>molecules and every mol of H2 Is 2 gram
40.34*10^23/6.22*10^23=6.48 or 6.5
6.5*2=13 gram
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4 0

3 years ago

Read 2 more answers

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide.

tatiyna

Answer:

$[H_2]_{eq}=0.183M$

$[I_2]_{eq}=0.183M$

$[HI]_{eq}=0.025M$

Explanation:

Hello.

In this case, for this equilibrium problem, we first realize that at the beginning there is just HI, it means that the reaction should be rewritten as follows:

$2HI\rightleftharpoons H_2+I_2$

Whereas the law of mass action (equilibrium expression) is:

$Kc=\frac{[H_2][I_2]}{[HI]^2}$

That in terms of initial concentrations and reaction extent or change $x$ turns out:

$Kc=\frac{x*x}{([HI]_0-2x)^2}\\\\54.3=\frac{x^2}{(0.391M-2x)^2}$

And the solution via solver or quadratic equation is:

$x_1=0.183M\\\\x_2=0.210M$

Whereas the correct answer is 0.183 M since the other value yield a negative concentration of HI at equilibrium (0.391-2*0.210=-0.029M).This, the equilibrium concentrations are:

$[H_2]_{eq}=0.183M$

$[I_2]_{eq}=0.183M$

$[HI]_{eq}=0.391M-2*0.183M=0.025M$