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skelet666 [1.2K]
3 years ago
14

Helppppppppppppppppppp

Chemistry
1 answer:
Lisa [10]3 years ago
3 0

Answer:

I I'm pretty sure it's B I hope this helps

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How many moles of water are produced from 373 mol Al??
stepladder [879]
I cannot come up with a reaction which you can convert directly from Al to H2O.

But you can convert from Al2O3 to H2O by adding HCL solvent.
5 0
3 years ago
Read 2 more answers
What happens to carbon in plants when the plants die.
Korvikt [17]

Answer:

Released into the atmosphere

Explanation:

The carbon that the plant had been storing when it was alive, would be released upon death into the soil

4 0
2 years ago
For the aqueous reaction dihydroxyacetone phosphate is the reactant and glyceraldehyde 3 phosphate is the product. dihydroxyacet
lbvjy [14]

<u>Answer:</u> The Gibbs free energy of the reaction is -445 J/mol.

<u>Explanation:</u>

The chemical equation for the conversion follows:

\text{Dihydroxyacetone phosphate}\rightleftharpoons \text{Glyceraldehyde-3-phosphate}

The expression for K_{eq} of above equation is:

K_{eq}=\frac{\text{[Glyceraldehyde-3-phosphate]}}{\text{[Dihydroxyacetone phosphate]}}

We are given:

[Glyceraldehyde-3-phosphate] = 0.00400 M

[Dihydroxyacetone phosphate] = 0.100 M

Putting values in above equation, we get:

K_{eq}=\frac{0.004}{0.100}=0.04

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G=\Delta G^o+RT\ln K_1

where,

\Delta G^o = Standard Gibbs free energy = 7.53 kJ/mol = 7530 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = temperature = 298 K

Putting values in above equation, we get:

\Delta G=7530J/mol+(8.3145J/Kmol)\times 298K\times \ln (0.04)\\\\\Delta G=-445J/mol

Hence, the Gibbs free energy of the reaction is -445 J/mol.

3 0
4 years ago
Two beakers are placed in a small closed container at 25 °C. One contains 284 mL of a 0.296 M aqueous solution of C6H12O6; the s
Katen [24]

Answer:

Final volume = 0.103M x446ml/0.184m = 250ml

Explanation:

As time passes, the volume of solutions in the second beaker decreases and that in the first beaker increases. If we wait long enough, the final volumes and concentration in the beakers would be,

First beaker

Final concentration = 0.184M

Final volume = 0.296M x 284ml/0.184 = 457ml

Second beaker

Final concentration = 0.184M

Final volume = 0.103M x446ml/0.184m = 250ml

6 0
3 years ago
Read 2 more answers
Complete and balance the following equation: S(s)+HNO3(aq)→H2SO3(aq)+N2O(g)(acidic solution)
meriva

Answer:- The balanced equation is, 2HNO_3(aq)+2S(s)+H_2O(l)\rightarrow N_2O(g)+2H_2SO_3(aq) .

Solution:- Oxidation number of S is increasing from 0 to 4 and so it is oxidation. Oxidation number of N is decreasing from 5 to 1 and so it is reduction.

We write the oxidation and reduction half equations and balance them. The given reaction is taking place in an acidic medium.

First of all we balance all the atoms other than H and O. Then oxygen is balanced by adding H_2O and hydrogen is balanced by adding H^+ . Charge is balanced by adding electrons.

To makes the electrons equal for both the half equations we multiply the equation/equations by appropriate numbers.

Oxidation half equation:

S(s)\rightarrow H_2SO_3(aq)

S is already balanced. To balance O, we need to add three water molecules to the left side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)

For balancing hydrogen, we need to add 4 hydrogen ions to the right side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)+4H^+(aq)

Now to balance the charge we need to add 4 electrons to the right side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)+4H^+(aq)+4e^-

Reduction half equation:

HNO_3(aq)\rightarrow N_2O(g)

To balance N, we need to multiply left side by 2:

2HNO_3(aq)\rightarrow N_2O(g)

For balancing oxygen, we need to add 5 water molecules to the right side:

2HNO_3(aq)\rightarrow N_2O(g)+5H_2O(l)

To balance hydrogen, we need to add 8 hydrogen ions to the left side:

2HNO_3(aq)+8H^+(aq)\rightarrow N_2O(g)+5H_2O(l)

Now, for charge balance, we need to add 8 electrons to the left side:

2HNO_3(aq)+8H^+(aq)+8e^-\rightarrow N_2O(g)+5H_2O(l)

First half equation has 4 electrons and second half equation has 8 electrons.

To make the electrons equal, we need to multiply oxidation half equation by 2:

2S(s)+6H_2O(l)\rightarrow 2H_2SO_3(aq)+8H^+(aq)+8e^-

Now we add both of these two half equations and cancel out common species. What we get on doing this is:

2HNO_3(aq)+2S(s)+H_2O(l)\rightarrow N_2O(g)+2H_2SO_3(aq)



6 0
3 years ago
Read 2 more answers
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