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Rainbow [258]
3 years ago
12

In the chemical reaction: NaOH (s) + H2O Na+ (aq) + OH- (aq), which substance (or substances) is in solution? A. Na+ (aq) only B

. OH- (aq) only C. Na+ (aq) and OH- (aq) D. NaOH (s)
The answer is C.
Chemistry
1 answer:
bonufazy [111]3 years ago
3 0

Answer:

in a chemical reaction of NaOH with H2O, after NaOH is completely disassociated, we will find Na+ and OH- ions in the solution. (option C).

Explanation:

In a reaction where NaOH is added to H2O.

NaOH is considered a strong base, this means that in an aqueous solution ( in water) it's  able to completely disassociate in ions.

There will not remain any NaOH in the solution. This means option D is not correct.

The ions in which NaOH will disassociate are : NaOH → Na+ + OH-

These ions we will find in the solution.

Not only Na+ because NaOH is a strong base, so there will be a lot of OH- ions as well in solution.

This means in a chemical reaction of NaOH with H2O, after NaOH is completely disassociated, we will find Na+ and OH- ions in the solution.

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Consider the following balanced equation:3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)If 24.2 moles of Na3PO4(aq) reac
Tomtit [17]

Answer:

69.7% is percent yield

Explanation:

Based on the reaction:

3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)

2 moles of Na3PO4 react producing 6 moles of NaNO3.

As 24.2 moles of Na3PO4 react, theoretical moles of NaNO3 produced are:

24.2 moles Na3PO4 * (6 moles NaNO3 / 2 moles Na3PO4) =

72.6 moles of NaNO3

As there are produced 50.6 moles of NaNO3, percent yield is:

50.6 moles NaNO3 / 72.6 moles NaNO3 =

<h3>69.7% is percent yield</h3>
7 0
3 years ago
When a sample of ca(s) loses 1 mole of electrons in a reaction with a sample of o2(g) the oxygen?
Stolb23 [73]
The reaction involved in present case is:

Net Reaction: Ca   +    1/2 O2    →         CaO. ..................(1)

In terms of oxidation and reduction, the reaction can be shown at

Oxidation: Ca     →       Ca2+      +       2e- .................(2)
Reduction: 1/2O2    +   2e-      →      O2-...................(3)

From, reaction 1 it can be seen that 1 mol of Ca reacts with 1/2 mol of O2 to form 1 mol of CaO.

From, reaction 2 it can be seen that 1 mol of Ca, generates 2 mol of e-.

Thus, when 1/2 mol of Ca is used in reaction, it will lose 1 mol of electrons.
6 0
3 years ago
Read 2 more answers
Suppose NAD is unavailable because NADH cannot be oxidized due to a mutation in the NADH dehydrogenase (Complex I). If FAD could
Dovator [93]

Answer:

FAD substitution will produce 28 ATP instead of 36.

Explanation:

NAD and FAD are coenzymes involved in reversible oxidation and reduction reactions. These compounds are also known as electron carriers. However NADH produce 3 electrons in electron transport chain and FADH2 produce 2 electron beacuase it transfer the electrons to second complex in ETC.

Normal prduction of ATP from glucose;

2 cytoplasmic NADH formed in glycolysis         Each yields 2 ATP   +4

2 NADH formed in the oxidation of pyruvate Each yields 3 ATP         +6

2 FADH2 formed in the citric acid cycle         Each yields 2 ATP         +4

6 NADH formed in the citric acid cycle             Each yields 3 ATP         +18

2 ATP from glycolysis                                                                                   +2  

2 ATP from citric acid cycle                                                                          +2                            

                                                                    Net yield ATP +36

C6H12O6 + 6 CO2 + 36 ADP + 36 Pi ⇒6 CO2 + 6 H2O + 36 ATP

If we replace the NAD with FAD the total ATP production would be.

2 cytoplasmic FADH2 formed in glycolysis          Each yields 2 ATP            +4

2 FADH2 formed in the oxidation of pyruvate   Each yields 3 ATP    +4

2 FADH2 formed in the citric acid cycle            Each yields 2 ATP    +4

6 FADH2 formed in the citric acid cycle                  Each yields 2 ATP   +12

2 ATP from glycolysis                                                                                    

+2

2 ATP from citric acid cycle                                                                            +2

                                                                  <u>Net yield ATP +28</u>

C6H12O6 + 6 CO2 + 28ADP + 28 Pi ⇒6 CO2 + 6 H2O + 28 ATP

7 0
3 years ago
Classify this chemical reaction:<br><br> 2 HNO3 (aq) + Sr(OH)2 (aq) → Sr(NO3)2 (aq) +2 H2O (l)
ad-work [718]

Answer:

This is not an answer its just to get points

7 0
3 years ago
A student heats 10.52 g of sodium hydrogen carbonate in a crucible until the compound completely decomposes to sodium carbonate
saveliy_v [14]

Answer:

m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3

Explanation:

Hello!

In this case, since the decomposition of sodium hydrogen carbonate is:

2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)

Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:

m_{Na_2CO_3}^{theoretical}=10.52gNaHCO_3*\frac{1molNaHCO_3}{84.01gNaHCO_3}*\frac{1molNa_2CO_3}{2molNaHCO_3}  *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}\\\\ m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3

Best regards!

4 0
3 years ago
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