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3 years ago

Urgent!!

Chemistry

2 answers:

madam [21]3 years ago

6 0

Answer:

Harmony is correct, because Mendeleev’s model made predictions that came true.

Explanation:

Mendeleev published periodic table.

Mendeleev also arranged the elements known at the time in order of relative atomic mass, but he did some other things that made his table much more successful.

Our answer is : Harmony is correct, because Mendeleev’s model made predictions that came true.

7nadin3 [17]3 years ago

4 0

Answer:

Harmony is correct, because Mendeleev’s model made predictions that came true.

Explanation:

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Increase food supplies in an ecosystem decrease competition because the competing organsims would

irga5000 [103]

The competing organisms would eventually outnumber the competition

6 0

3 years ago

1.25 is the closest to 1.04 or not I want to answer please. I think it's true, but I want to prove it scientifically, please.

PSYCHO15rus [73]

Answer:

false because if you round both of them

3 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

4 years ago

How many molecules are in a 2.1 mol of H2O? *

gayaneshka [121]

Answer:

The mole is the SI unit for amount of a substance.

Explanation:

Just like the dozen and the gross, it is a name that stands for a number. There are therefore 6.02 × 10 23 water molecules in a mole of water molecules.

6 0

3 years ago

Answer the following questions…

Zina [86]

The concentration of the hydronium ion in hydrochloric acid is 0.0045 M, and the pH of the solution is 2.34.

pH is the potential of the hydrogen or the hydronium ions in the aqueous solution.

As the solution contains $4.5 \times 10^{-3} \;\rm M\;$ HCl the concentration of the hydronium ion will be the same, $4.5 \times 10^{-3} \;\rm M.$

The pH of the solution is calculated as:

$\begin{aligned} \rm pH &= \rm -log[H^{+}]\\\\&= - \rm log (4.5 \times 10^{-3})\\\\&= 2.34\end{aligned}$

The concentration of the hydroxide ion is calculated from pH and hydronium ion as:

$\begin{aligned} \rm [H_{3}O^{+}][OH^{-}] &= 10^{-14}\\\\&= \dfrac{1 \times 10^{-14}}{4.5 \times 10^{-3}}\\\\&= 2.2 \times 10^{12}\end{aligned}$

Now, for the calcium hydroxide solution, the calculations are shown as,

$\begin{aligned} \rm (H_3}\rm O^{+}) &= \rm antilog (-pH)\\\\&= \rm antilog (-8)\\\\&= 10^{-8} \;\rm M\end{aligned}$

pOH is calculated as:

$\begin{aligned} \rm pOH &= 14- 8 = 6\\\\\rm [OH^{-}] &= \rm antilog (-6)\\\\&= 10^{-6} \end{aligned}$

The concentration of calcium hydroxide is calculated as:

$\begin{aligned} &= \dfrac{1}{2} \times \rm [OH^{-}]\\\\&= 5 \times 10^{-4} \;\rm M\end{aligned}$

Therefore, the pH and the pOH give the concentration of the hydrogen or the hydronium ion and the hydroxide ion.

Learn more about pH and pOH here:

brainly.com/question/16062632

#SPJ1

3 0

2 years ago