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3 years ago

11

Help me i am on a quiz

2 answers:

castortr0y [4]3 years ago

6 0

The screen is blank for me can you say the question

Sladkaya [172]3 years ago

4 0

Answer: The rock on the right is foliated.

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Help out this a quick one

Gemiola [76]

Answer:

Climate

Explanation:

The 3 pieces of evidence are fossils, rocks, and mountain ranges all are included except climate

6 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago

A chemist titrates 110.0 mL of a 0.7684 M methylamine (CH3NH2) sotion with 0.4469 M HNO3 solution at 25 °C. Calculate the pH at

Firdavs [7]

Answer: The pH at equivalence point for the given solution is 5.59.

Explanation:

At the equivalence point,

$n_{HNO_{3}} = n_{CH_{3}NH_{2}}$

So, first we will calculate the moles of $CH_{3}NH_{2}$ as follows.

$n_{CH_{3}NH_{2}} = 0.764 M \times \frac{110 ml}{1000 ml/L}$

= 0.0845 mol

Now, volume of $HNO_{3}$ present will be calculated as follows.

Volume = $\frac{\text{no. of moles}}{\text{Molarity}}$

= $\frac{0.0845}{0.4469 M}$

= 0.1891 L

Therefore, the total volume will be the sum of the given volumes as follows.

110 ml + 189.1 ml

= 299.13 ml

or, = 0.2991 L

Now, $[CH_{3}NH_{3}^{+}] = \frac{0.0845 mol}{0.2991 L}$

= 0.283 M

Chemical equation for this reaction is as follows.

$CH_{3}NH_{3}^{+} + H_{2}O \rightleftharpoons CH_{3}NH_{2} + H_{3}O^{+}$

As, $k_{a} = \frac{k_{w}}{k_{b}}$

= $\frac{10^{-14}}{10^{-3.36}}$

= $2.29 \times 10^{-11}$

Now, $[HNO_{3}] = \sqrt{k_{a}[CH_{3}NH_{3}^{+}]}$

= $\sqrt{2.29 \times 10^{-11} \times 0.283}$

= $2.546 \times 10^{-6}$

Now, pH will be calculated as follows.

pH = $-log [H_{3}O^{+}]$

= $-log (2.546 \times 10^{-6})$

= 5.59

Thus, we can conclude that pH at equivalence point for the given solution is 5.59.

6 0

3 years ago

The transition metals are located in what block of the periodic table

gladu [14]

Located in the D block

7 0

3 years ago

A molecular compound is?

Nezavi [6.7K]

Composed of molecules formed by atoms of two or more different elements.

3 0

4 years ago

Read 2 more answers