Ask question

Ask question

All categories

3 years ago

10

Write the precipitation reaction for chromium(III) nitrate in aqueous solution: Use the pull-down menus to specify the state of

each reactant and product. Is chromium(III) nitrate considered soluble or not soluble

1 answer:

Licemer1 [7]3 years ago

6 0

Answer:

Nitrate is considered SOLUBLE

Explanation:

please kindly refer to attachment for the step by step solution of the given problem.

You might be interested in

Please answer ASAP,

andrew11 [14]

Answer:

B

Explanation:

I took the test and GOT an A

6 0

3 years ago

why is the frequency of the C=O stretch in the hydrogenation product higher than in ethyl cinnamate?

antiseptic1488 [7]

The hydrogenation of ethyl cinnamate will result to an ester which can be a racemic mixture of products or just a meso compound depending on the structure of the ethyl cinnamate (cis or trans). The ester will produce a frequency of C=O stretch of <span>1750 - 1735 (s.</span>

7 0

3 years ago

Kahdbiaddsk;wvfodas vs;oi vsd;iv sdv;sidv s;vowdbvw;dichvbwdv;iwebfwd;iwdbcv;wkfbsdliasbv;soqivbasd

storchak [24]

Answer:

I am not positive sorry

6 0

3 years ago

Read 2 more answers

All faculty members are happy to see students help each other. Dumbledore is particularly pleased with Hermione. Though, it shou

____ [38]

Answer:

$m_{Mg}=30.8mgMg$

Explanation:

Hello,

Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:

$n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2$

Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:

$m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg$

Best regards.

3 0

3 years ago

Determine the concentration of nh3(aq) that is required to dissolve 743 mg of agcl(s) in 100.0 ml of solution. the ksp of agcl i

AnnyKZ [126]

Answer:

1.1 M

Explanation:

The dissociation of $AgCl_{(s)}$ is as follows:

$AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}$

Given Value for $K_{sp} = 1.77*10^{-10}$

The equation for the reaction for the formation of complex ion $Ag(NH_3)^+_2$ is :

$Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}$

The value of $K_f = 1.6*10^7$

If we combine both equation and find the overall equilibrium constant will be:

$AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}$

$Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}$

<u> </u>

$AgCl_{(s)}+2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}$

$K = (1.77*10^{-10})(1.6*10^7)$

$K = 0.00283$

If $[NH_3]$ = x M

The solubility of $AgCl_{(s)}$ in the $NH_3$ solution will be:

$x = 743*10^{-3}g * \frac{mol AgCl}{143.32g}*\frac{1}{0.1000L}$

$x = 0.0518 M$

Constructing an ICE Table; we have :

$AgCl_{(s)} + 2 NH_3{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}$

Initial (M) x 0 0

Change (M) -2 (0.0518) + 0.0518 + 0.0518

Equilibrium (M) x - 0.1156 0.0518 0.0518

Equilibrium constant;

(K) = $\frac{[Ag(NH_3)_2^+][Cl^-]}{[NH_3]^2}$

$0.00283 = \frac{(0.0518)^2}{(x-0.1156)^2}$

$0.00283 = (\frac{0.0518}{x-0.1156})^2$

$x = 0.1156 + \sqrt{\frac{0.0518^2}{0.00283} }$

x = [NH₃] = 1.089 M

[NH₃] ≅ 1.1 M

8 0

3 years ago