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Olin [163]
4 years ago
8

A quantity of CO gas occupies a volume of 0.32 L at 0.90 atm and 323 K . The pressure of the gas is lowered and its temperature

is raised until its volume is 3.8 L . Find the density of the CO under the new conditions. Express your answer to two significant figures and include the appropriate units.
Chemistry
1 answer:
Contact [7]4 years ago
5 0
Mathematical formula of Ideal Gas Law is PV=nRT
  where: P-pressure, 
              V-volume
              n-number of moles; m/MW
              T-Temperature
              m-mass
              d-density ; m/V
              MW-Molecular Weight
              R- Ideal Gas constant. If the units of P,V,n & T are atm, L, mol & K respectively, the value of R is 0.0821 L x atm / K x mol
    
Substituting the definitions to the original Gas equation becomes:
     d= P x MW / (RxT)

Solution : d= .90atm x 28 g/mol (CO) / 0.0821Lxatm / mol x K  x 323 K
           
                d = 25.2 g / 26000 mL

                d = .0.00096 g/mL is the density of CO under the new conditions
           
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Iteru [2.4K]
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7 0
3 years ago
1. How many grams of B are present in 3.35 grams of boron tribromide ?
Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

                            251 g of BBr₃ ----------------------  11 g of B

                             3.35 g          -----------------------    x

                           x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

                             251g of BBr₃ -----------------   80 g of Br

                                   x               ----------------- 4.69 g

                           x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

________grams N.

MW NF₃ = 71 g

                               71 g of NF₃    -----------------   14 g of N

                               4.11 g              ----------------     x

                               x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

MW NF₃ = 71

                                71 g of NF₃ ---------------------   19 g of F

                                  x               ---------------------   3.07 g

                                 x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

                             440 g of CoI₃ ------------------  59 g of Co

                              1.16 g             ------------------   x

                              x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

________grams cobalt(III) iodide.

MW CoI₃ = 440 g

                             440 g of CoI₃ ------------------  59 g of Co

                               x                   -----------------   2.28 g of Co⁺³

                              x = (2.28 x 440) / 59

                              x = 17 g of CoI₃

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3 years ago
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Hunter-Best [27]
Co2 is correct buddy
7 0
4 years ago
How many grams are in 2.94 x 1017 atoms of Iron?
ale4655 [162]

Answer:

8.677

⋅

10

8

g

Explanation:

5 0
3 years ago
A system does 125 J of work and cools down by releasing 438 J of heat. The change in internal energy is ____________ J.
Jet001 [13]

Given that,

Work done by the system = 125 J

Energy released when it cools down = 438 J

To find,

The change in internal energy.

Solution,

As heat is released by the system, Q = -438 J

Work done by the system, W = -125 J

Using the first law of thermodynamics. The change in internal energy is given by :

\Delta U=Q-W\\\\=-438-125\\\\=-563\ J

So, the change in internal energy is 563 J.

8 0
3 years ago
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