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butalik [34]
3 years ago
5

A sample of 0.010 mole of oxygen gas is confined at 127 °c and 0.80 atmosphere. what would be the pressure of this sample at 27

°c and the same volume?
Chemistry
1 answer:
devlian [24]3 years ago
4 0
First,let's assume ideal gas behavior for simplicity. This is a special case because the volumes of the two states are equal. At constant volume, we can use the Gay-Lussac equation:

P₁/T₁ = P₂/T₂

(0.8)/(127+273) = (P₂)/(27+273)
Solving for P₂,
P₂ = 0.6 atm

<em>Thus, the answer is 0.6 atm.</em>
You might be interested in
What is the percent mass oxygen in calcium carbonate (CaCo3)?
Gnoma [55]
  <span>Step 1 is to determine the mass of each part 
Mass of Ca is 40.08 g 
Mass of C is 12.01 g 
Mass of O is 16.00 x 3 = 48.00 g 
Step 2 is to determine the total mass of the compound 
Total mass of CaCO3 is 40.08 + 12.01 + 48.00 = 100.09 g 

Step 3 is to determine the % of each part using the following formula: 
Mass of part / total mass x 100 = 

40.08 / 100.09 x 100 = 40.04 % Ca 

12.01 / 100.09 x 100 = 12.00 % C 

48.00 / 100.09 x 100 = 47.96 % O 

Step 4 is to double check by adding all percentages. If they equal 100, then I probably did it right. :) 
40.04 
+12.00 
+47.96 
=100.00</span><span>
</span>
4 0
3 years ago
Read 2 more answers
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
Consider the following ionic compounds: CdCO3, Na2S, PbSO4, (NH4)3PO4, and Hg2Cl2. Which compounds will be soluble when added to
k0ka [10]

Answer:

Na2S, (NH4)3PO4

Explanation:

We can decide what compounds are soluble by considering the solubility rules that apply.

CdCO3 is not soluble in water because all carbonates are insoluble except those of ammonium, sodium and potassium.

Na2S is soluble in water because all sulphides are insoluble except those of sodium, potassium and ammonium.

PbSO4 is insoluble in water because all sulphates are soluble except those of lead and barium. The sulphate of calcium is only slightly soluble in water.

(NH4)3PO4 is soluble in water because all phosphates are insoluble except those of sodium, potassium and ammonium.

Hg2Cl2 is insoluble in water because all chlorides are soluble except those of lead, mercury II and silver.

7 0
3 years ago
Astudent prepareda calibration curve by plotting absorbance of the standards against the [FeSCN2+] molar concentration (M). The
Nuetrik [128]

Answer:

Explanation:

Chemistry 1B Experiment 7

1-3 5.0 1.5 3.5

Part 2: Determining the equilibrium constant.

Label 5 medium-sized test tubes. Table 7.2 shows the amounts of 2.00 × 10–3

M

Fe(NO3)3 (in 1 M HNO3) solution, 2.00 × 10–3

M KSCN solution, and purified water

that should be added to each tube. Pipet the approximate amount of each solution into

each tube. (Record the exact amount of each solution that you actually add. You will

need to use these actual amounts in your calculations.)

Obtain five separate small pieces of parafilm. Close the top of each test tube with

the parafilm. Mix each solution thoroughly by inverting the test tube several times.

Record your observations.

Measure and record the absorbance of each solution at the 447 nm.

Table 7.2 Composition of solutions for determining the equilibrium constant.

Test Tube

Volume of

2.00 × 10–3

M Fe(NO3)3

in 1 M HNO3 (mL)

Volume of

2.00 × 10–3

M KSCN

(mL)

Volume of

purified water

(mL)

2-1 5.0 1.0 4.0

2-2 5.0 2.0 3.0

2-3 5.0 3.0 2.0

2-4 5.0 4.0 1.0

2-5 5.0 5.0 none

Calculations

Part 1. Graphing the relationship between absorbance and [FeSCN2+].

Assuming that “all” of the SCN–

ions have been converted to FeSCN2+ ions,

calculate [FeSCN2+] in each of the solutions in Part 1. For example, in test tube 1-2, 1.0

mL of a 2.00 × 10–3

M KSCN solution was diluted to 10.0 mL. The concentration of

SCN–

that results from this dilution is the one to use for determining [FeSCN2+].

Because of the 1:1 stoichiometry, that initial concentration of SCN– is equal to

[FeSCN2+].

Plot a full-page graph of the absorbance against the concentration of FeSCN2+ in

all standard solutions. Use a ruler to draw the best straight line that comes closest to each

of your five data points. Your line should pass through (0 M, 0). (Why?) This graph is

your calibration curve. When you measure the absorbance of a solution that contains an

unknown concentration of FeSCN2+ ions, you can use this calibration curve to determine

the unknown concentration

6 0
4 years ago
What are the oxidation numbers of the atoms in this reaction?
OLEGan [10]
<span>
all these are correct:
0 for Mg(s) and H in H2

</span><span>+1 for H in HCl

</span><span>-1 for Cl in HCl
</span>
<span>+2 for Mg in MgCl2

only the first statement is wrong</span>
5 0
3 years ago
Read 2 more answers
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