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Reika [66]
3 years ago
10

A student increases the temperature of a 556 cm3 balloon from 278 K to 308 K. Assuming constant pressure, what should the new vo

lume of the balloon be? A --376cm3
B --462cm3
C --924cm3
D --417cm3
Chemistry
1 answer:
Mandarinka [93]3 years ago
3 0
The answer is:  [D]:  " 417 cm³ " .
_____________________________________________________
Explanation:  Use the formula:

V₁ /T₁= V₂ /T₂  ;

in which:  V₁ = initial volume = 556 cm³ ;
                T₁ = initial temperature = 278 K ;
                V₂ = final ("new") temperature = 308 K
                T₂ = final ("new:) volume = ?

Solve for  "V₂" ;

Since:  V₁ /T₁= V₂ /T₂ ;

We can rearrange this "equation/formula" to isolate "V₂" on one side of the equation; and then we can plug in our know values to solve for "V₂" ;
_______________________________________________________
       V₁ /T₁= V₂ /T₂  ;  Multiply EACH side of the equation by "T₂ " :

          →  T₂ (V₁ /T₁) = T₂  (V₂ /T₂) ;
______________________________
to get:

↔ T₂  (V₂ /T₂) = T₂ (V₁ /T₁) ;

     →  V₂ = T₂ (V₁ /T₁) ;
______________________________
Now, plug in our known values, to solve for "V₂" ;
______________________________
    →  V₂ = T₂ (V₁ /T₁) ;
______________________________
    →  V₂ = 308 K ( 556 cm³ /278 K)  ;
             → The units of "K" cancel to "1" ; and we have:
________________________________________________________
    →  V₂ = 308*( 556 cm³ / 278 ) = [(208 * 556) / 278 ] cm³ ;
Note:  We will keep the units of volume as:  "cm³ ";  since all the answer choices given are in units of:  "cm³ " ; {that is, "cubic centimeters"}.

   →  [(208 * 556) / 278 ] cm³ = [ (115,648) / (278) ] cm³ ;
                                        
              → For the "(115,648)" ;  round to "3 (three significant figures)" ;
                        → "(115,648)" → rounds to:  "116,000" ;
____________________________________________________
              →      (116,000) / (278) = 417.2661870503597122  ;
                                 → round to 3 significant figures; → "417 cm³ " ;
                                               → which corresponds with "choice [D]".
______________________________________________________
The answer is:  [D]:  "417 cm³ " .
______________________________________________________

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Tropic levels have most energy
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The hot dogs you ate at the barbecue last week were 75% fat-free by weight, had 275 calories, and weighed 110 g. a) What percent
vovangra [49]

Answer:

a) 25%

b) 27.5 g

c) 90%

Explanation:

a) 75% fat-free by weight means 25% of the weight is made by fat.

b) 110 g ___ 100%

      x    ___ 25%

          x = 27.5g

Each hot dog has 27.5g of fat.

c) 9 cal ___ 1 g fat

      y    ___ 27.5 g fat

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275 cal ___ 100%

247.5 cal ___ z

     z = 90%

90 % of the calories come from fat.

6 0
3 years ago
A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new
Artyom0805 [142]

Answer: There are now 2.07 moles of gas in the flask.

Explanation:

PV=nRT

P= Pressure of the gas = 697 mmHg = 0.92 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = ?

n = number of moles = 1.9

T = Temperature of the gas = 21°C=(21+273)K= 294 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

V=\frac{nRT}{P}=\frac{1.9\times 0.0821 \times 294}{0.92}=49.8L

When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

PV=nRT

P= Pressure of the gas = 775 mmHg = 1.02 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

n=\frac{PV}{RT}=\frac{1.02\times 49.8}{0.0821\times 299}=2.07moles

Thus the now the container contains 2.07 moles.

6 0
2 years ago
If a dog has a mass of 16.1 kg, what is its mass in the following units? Use scientific notation in all of your answers.
nekit [7.7K]

Answer:

16.1\times 10^3\ \text{g}

16.1\times 10^{6}\ \text{mg}

16.1\times 10^{9}\ \mu\text{g}

Explanation:

Mass of dog = 16.1\ \text{kg}

1\ \text{kg}=10^{3}\ \text{g}

1\ \text{kg}=10^{6}\ \text{mg}

1\ \text{kg}=10^{9}\ \mu\text{g}

So,

16.1\ \text{kg}=16.1\times 10^3\ \text{g}

16.1\ \text{kg}=16.1\times 10^{6}\ \text{mg}

16.1\ \text{kg}=16.1\times 10^{9}\ \mu\text{g}

Mass of the dog in the other units is 16.1\times 10^3\ \text{g}, 16.1\times 10^{6}\ \text{mg} and 16.1\times 10^{9}\ \mu\text{g}.

3 0
2 years ago
A 2.10 L vessel contains 4.65 mol of nitrous oxide (N2O) at 3.82 atm. What will be the pressure of this gas in a container that’
sveticcg [70]

Answer:

The answer to your question is 7.64 atm

Explanation:

Data

Volume 1 = V1 = 2.1 l

moles = 4.65

Pressure 1 = P1 = 3.82 atm

Volume 2 = Volume 1/2

Pressure 2 = ?

Process

1.- Calculate the new volume

Volume 2 = 2.10/2

Volume 2 = 1.05 l

2.- Use Boyle's law to find the Pressure

             P1V1 = P2V2

-Solve for P2

              P2 = P1V1 / V2

-Substitution

              P2 = (3.82)(2.1) / 1.05

-Simplification

              P2 = 8.022/1.05

-Result

              P2 = 7.64 atm

4 0
3 years ago
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