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3 years ago

Three most common alkali metals include:

Chemistry

2 answers:

QveST [7]3 years ago

4 0

Answer:

Sodium, potassium, and rubidium

Explanation:

The relative abundances in the Earth’s crust are:

Na = 2.6 %; K = 2.4 %; Rb = 60 ppm; Li = 17 ppm; Cs = 3 ppm; Fr ≈ 0 ppb

Leona [35]3 years ago

3 0

Sodium, Potassium, and Cesium

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The 38 elements in groups 3 through 12 of the periodic table are called "Alkaline Earth Metals" O True O False

kotykmax [81]

Answer:

FALSE

they're called transition metals

5 0

3 years ago

Read 2 more answers

If you start with 6 mol of nitrogen gas (N2+) what mass (g) of ammonia (NH4) will be produced?

Yanka [14]

Answer:

204g of NH3

Explanation:

The balanced equation for the reaction is given below:

N2 + 3H2 —> 2NH3

Next, we shall determine the number of mole NH3 produced by reacting 6moles of N2. This is illustrated below:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 6 moles of N2 will react to produce = 6 x 2 = 12 moles of NH3.

Finally, we shall convert 12 moles of NH3 to grams. This is illustrated below:

Number of mole of NH3 = 12 moles.

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 =..?

Mass = mole x molar mass

Mass of NH3 = 12 x 17

Mass of NH3 = 204g.

Therefore, 204g of NH3 will be produced from the reaction.

3 0

3 years ago

According to Le Châtelier's principle, a system at equilibrium will respond to a stress by shifting in the direction that reliev

Romashka [77]

Answer:

a. reduces the stress

Explanation:

3 0

4 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 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

3 years ago