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3 years ago

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Which of the following processes involves an oxidation-reduction reaction and allows engineers to coat one metal with another? 1

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1 answer:

FrozenT [24]3 years ago

7 0

Answer: The correct answer is C.

Explanation: Hope this helps plz mark brainliest.

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(40 Points) Complete, balance, compute the amounts of the products assuming 100% yield. 10g Na + 10g Oxygen.

Lilit [14]

13.5g

Explanation:

Given parameters:

Mass of Na = 10g

Mass of O₂ = 10g

Unknown:

Mass of products formed = ?

Balanced equation = ?

Solution:

The balanced chemical equation is shown below:

4Na + O₂ ⇒ 2Na₂O

In any reaction, the specie in short supply determines the extent of the reaction.

This reaction is not an exclusion. We need to first determine the specie in short supply and use it to estimate the amount of product since we have a 100% yield which signifies that all was used up.

let us convert to moles;

Number of moles of Na = $\frac{mass }{molar mass} = \frac{10}{23}$ = 0.435mole

Number of moles of O₂ = $\frac{mass}{molar mass} = \frac{10}{32}$ = 0.313mole

From the given equation;

4 moles of Na requires 1 mole of O₂;

0.435 moles of Na will require $\frac{0.435}{4}$ = 0.11 moles

But the given amount O₂ is 0.313, this is an excess of 0.313 - 0.11 = 0.203moles

We see that Na is the limiting reagent;

4 moles of Na gives 2 mole of Na₂O

0.435 moles of Na will give $\frac{0.435 x 2 }{4}$ = 0.22 moles

Mass of Na₂O = number of moles x molar mass = 62 x 0.22 = 13.5g

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

5 0

4 years ago

Indicate whether the following balanced equations involve oxidation-reduction. Check all that apply. Check all that apply. 2H2SO

guapka [62]

Answer : The balanced equations involve oxidation-reduction are:

(a) $2H_2SO_4(aq)+2NaBr(s)\rightarrow Br_2(l)+SO_2(g)+Na_2SO_4(aq)+2H_2O(l)$

(b) $3SO_2(g)+2HNO_3(aq)+2H_2O(l)\rightarrow 3H_2SO_4(aq)+2NO(g)$

(c) $NaI(aq)+3HOCl(aq)\rightarrow NaIO_3(aq)+3HCl(aq)$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(a) The given chemical reaction is:

$2H_2SO_4(aq)+2NaBr(s)\rightarrow Br_2(l)+SO_2(g)+Na_2SO_4(aq)+2H_2O(l)$

This reaction involve oxidation-reduction reaction because the oxidation state bromine changes from (-1) to (0) which shows oxidation and sulfur changes from (+6) to (+4) which shows reduction.

(b) The given chemical reaction is:

$3SO_2(g)+2HNO_3(aq)+2H_2O(l)\rightarrow 3H_2SO_4(aq)+2NO(g)$

This reaction involve oxidation-reduction reaction because the oxidation state sulfur changes from (+4) to (+6) which shows oxidation and nitrogen changes from (+5) to (+2) which shows reduction.

(c) The given chemical reaction is:

$NaI(aq)+3HOCl(aq)\rightarrow NaIO_3(aq)+3HCl(aq)$

This reaction involve oxidation-reduction reaction because the oxidation state iodine changes from (-1) to (+5) which shows oxidation and chlorine changes from (+5) to (-1) which shows reduction.

(d) The given chemical reaction is:

$PBr_3(l)+3H_2O(l)\rightarrow H_3PO_3(aq)+3HBr(aq)$

This reaction does not involve oxidation-reduction reaction because the oxidation state of element present on reactant and product side are same.

3 0

3 years ago

What is the molecular formula of a compound that contains 66% Ca and 34% P and

Lostsunrise [7]

Answer:

Ca3P2

Explanation: find out what 66% of the molar weight of the molecule is and then find out how many moles of Ca that would be by dividing it by the molar weight of Ca

then do the same for P

6 0

3 years ago

Assume that a firm has the following marginal benefit of pollution (denoted E for emissions, measured in tons): MB=150-5 E e. No

jeka57 [31]

Answer:

Explanation:

E)cost of pollution is reduction in benefit of the firms.

MB=150-5E,. MB=90-3E

E=30-MB/5. E=30-MB/3

Joint MB,

E=60-8MB/15

MB=112.5-1.875E

Total pollution reduction=24

Total pollution=60-24=36

MB=112.5-1.875*36=112.5-67.5=45

Firm 1

MB=150-5E.

45=150-5E.

E=-105/-5=21

Reduction=30-21=9

Firm 2,

MB=90-3E

45=90-3E

E=-45/-3=15

Reduction =30-15=15

So firm 2 is reducing 15 units of pollution and firm 1 reducing 9 units of pollution.

As each firm have to reduce 12 units of pollution but firm 2 reducing 3 units more while firm 1 reducing 3 less units.

So, firm 2 is selling 3 units of pollution emission permit to firm 1.

F)firm 1 reduce 9 units and firm 2 reduce 12 units of pollution after trade.

Total cost of pollution reduction will total Benefit reduction by pollution reduction.

MB=112.5 -1.875E

TB=112.5E-0.9375E^2

TB at E=60

TB=112.5*60-0.9375*60*60=6750-3375=3375

TB at E=36

TB=112.5*36-0.9375*36*36=4050-1215=2836

Total cost of pollution reduction=3375-2836=540

G)price of permit= cost of extra pollution reduction by firm 2 or total cost from 9 to 12 by firm 2

MB=90-3E

TB=90E- 1.5E^2

TB at E=18

TB=90*18 -1.5*18*18=1620-486=1134

TB at E=15

TB=90*15 -1.5*15*15=1350-337.5=1012.5

Permit price=1134-1012.5=121.5

Total cost to firm 2 =1134

Net total cost to firm 2=1134-121.5=1012.5

Total cost to firm 1=150E-2.5E^2=150*9-2.5*9*9=1350-202.5=1147.5

Net total cost=1147.5+121.5=1269

H) the total cost is lower in cap & trade policy is because the firm who has higher cost of pollution reduction is paying the other firm who has lower cost of pollution reduction to reduce more pollution ,so that his part of pollution reduction can be completed.

And the amount he is paying is equal to the amount that is cost of other firm of reducing additional pollution units.

So the cost the firm is lower as he is paying lower amount than if he reduce pollution by itself.

5 0

4 years ago

Which of the following contains an atom that does not obey the octet rule? BrF3 SiO2 NaCl BrF

mrs_skeptik [129]

The octet rule signifies towards the capability of the atoms to prefer to exhibit eight electrons in the valence shell. In case, when the atoms possess lesser than eight electrons, they seem to react and produce more stable components. While discussing octet rule, one does not consider d or f electrons.

According to the octet rule, the atoms of the prime-group elements seem to bind in such a manner that each atom exhibit eight electrons in its valence shell, providing it the similar electronic configuration as a noble gas.

In the given question, in BrF3, fluorine follows octet rule, while bromine does not follow the octet rule.

5 0

3 years ago

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