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3 years ago

Can someone please explain the steps to the correct answers?

Chemistry

1 answer:

kondor19780726 [428]3 years ago

4 0

Explanation:

1. b is correct because 10^-2 is 1/100. multiplying that by 1.862 is 0.01862

2. 6 feet 10 inches = 82 inches. multiplying 82 by 2.54 we get 208.28. there are only two sig figs so it's 210 cm, not 0.21 cm.

3. 25.2 x 39370 = 992000, which is 9.92 x 10^5.

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Draw all the structural isomers for the molecular formula C3H8O. Be careful not to draw any structures by crossing one line over

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Answer:

The three isomers having the molecular formula $C_{3} H_{8}O$ are drawn in the figure below.

Explanation:

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3 years ago

Consider the reaction.

N76 [4]

Answer: m = 50 g ZnSO4

Explanation: First is convert the moles of Zn to the moles of ZnSO4 by having their mole ratio which is 2:2 based from the balanced equation. Next is convert the moles of ZnSO4 to mass using its molar mass.

0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn

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0.311 moles ZnSO4 x 161 g ZnSO4 / 1 mole ZnSO4

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4 years ago

Read 2 more answers

. 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

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4 years ago

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D. Sample C would be best

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3 years ago

Calculate the oxidation number of “C” in CH3COCH3 please

Triss [41]

Answer:

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Explanation:

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4 years ago