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kondor19780726 [428]
1 year ago
6

Suppose that on a hot and sticky afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab

(volume of 180 m3) was 1.1 atm before the storm and 0.85 atm during the storm, to what volume would the laboratory try to expand in order to make up for the large pressure difference outside?
190 m3
230 m3
1,800 m3
7,100 m3
Chemistry
2 answers:
uranmaximum [27]1 year ago
5 0

The answer is B) 230 m3

Andreyy891 year ago
5 0

Answer: 230m^3

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2

where,

P_1 = initial pressure of gas  = 1.1 atm

P_2 = final pressure of gas  = 0.85 atm

V_1 = initial volume of gas  = 180m^3

V_2 = final volume of gas  = ?

1.1\times 180=0.85\times V_2

V_2=230m^3

Therefore, the gas would expand to volume of 230m^3

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k = 23045 N/m

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To find the spring constant, you take into account the maximum elastic potential energy that the spring can support. The kinetic energy of the car must be, at least, equal to elastic potential energy of the spring when it is compressed to its limit. Then, you have:

K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2    (1)

M: mass of the car = 1050 kg

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v: velocity of the car = 8 km/h

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You solve the equation (1) for k. But first you convert the velocity v to m/s:

v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}

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1 year ago
How many grams of the excess reagent are left over when 6.00g of CS2 gas react with 10.0g of Cl2 gas in the following reaction:
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Answer:

CS₂ = 2.43 g

Explanation:

Data Given:

CS₂ gas = 6.00g

Cl₂ = 10.0g

Reaction Given:

                   CS₂(g) + 3Cl₂(g) --------> CCl₄(l) + S₂Cl₂(l)

Solution

Limiting Reagent :

The reactant which is less in amount control the amount of product and know as limiting reagent.

Excess Reagent:

The amount of reactant which are in excess and leftover at the end of reaction and product formed.

Now we have to find the reactant that is in excess

For this we will look at the Reaction

                       CS₂   +    3Cl₂  -------->   CCl₄   +  S₂Cl₂

                       1 mol     3 mol               1 mol      1 mol

we come to know from the above reaction that

1 mole of CS₂ react with 3 mole of Cl₂ to produce 1 mole of CCl₄ and 1 mole of  S₂Cl₂

Now to convert mole to mass we required molar masses

molar mass of CS₂ = 12 +  2(32)

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molar mass of Cl₂ = 2(35.5)

molar mass of Cl₂ = 71 g/mol

if we represent mole in grams then

             CS₂              +    3Cl₂             -------->   CCl₄   +  S₂Cl₂

      1 mol (76 g/mol)     3 mol (71 g/mol)

                CS₂   +    3Cl₂   -------->   CCl₄   +  S₂Cl₂

                 76 g       213 g

So,

we come to know that 76 g of CS₂ will combine with 213 g of Cl₂ to form product.

So now look for the ratio of both reactant

                   CS₂    :    Cl₂

                   76 g   :    213 g

                    1  g     :      2.8 g

So, the for every one gram of CS₂ required 2.8g Cl₂

So from this details

we apply unity formula

                 2.8 g of Cl₂ ≅  1 g of CS₂

                 10 g of Cl₂ ≅  ? g of CS₂

by doing cross multiplication

                    g of CS₂ = 10 x 1 / 2.8

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10.0g of Cl₂ used 3.57 g of CS₂

It showed that 3.57 g of CS₂ used and the remaining amount of CS₂ is

               Remaining amount of CS₂ = 6 - 3.57 = 2.43 g

 

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