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Sholpan [36]
4 years ago
10

Wht is 30.00 grams of H3PO4

Chemistry
1 answer:
LenaWriter [7]4 years ago
3 0

Answer:

0.31 mol

Explanation:

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Darker colors absorb sunlight more, so it will become hotter. Lighter colored clothes will better reflect sun so they will be less hot.

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7 0
3 years ago
Which solution has the highest boiling point at standard pressure?
julia-pushkina [17]
Pressure: when the external pressure is:
less than one atmosphere, the boiling point of the liquid is lower than its normal boiling point.
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8 0
3 years ago
Read 2 more answers
Please help me as soon as possible! Please help me as soon as possible!
Nastasia [14]

The Answer is D. Suspending a heavy weight with a strong chain.

6 0
4 years ago
Hi can u help me pls? I'm totally stuck . The natural source of acidity in rain water is _____.
kykrilka [37]

Answer-The correct option is option d with says all of the above.

Explanation- All three acids that are given combined together to form acid rain in which nitric and sulphuric acid are stronger acids present while carbonic acid is a weaker one.

The carbon dioxide admitted in air combines with water to form carbonic acid and gives a weak acidic nature to rainwater. Pollution in nature makes sulphur and nitrogen present in air react to form the stronger acids responsible for acid rain.

5 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
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