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Alex
2 years ago
5

If the volume of a container of gas is reduced, what will happen to the pressure inside the container?

Chemistry
2 answers:
blondinia [14]2 years ago
8 0
B).The pressure will not change

andrew11 [14]2 years ago
5 0

Answer: a. The pressure will increase.

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

P_2 = final pressure of gas

V_1 = initial volume of gas

V_2 = final volume of gas

Thus as volume of the container will be reduced, the pressure would increase.

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Phosphorus has the molecular formula p4, and sulfur has the molecular formula s8. how many grams of phosphorus contain the same
Alinara [238K]
1) Find the number of molecules in 7.88 g of sulfur

molar mass of S8 = 8*atomic mass of S = 8 * 32.0 g / mol = 256.0 g/mol

Number of moles  = mass in grams / atomic mass = 7.88 g / 256.0 g / mol = 0.0308 moles

2) Find the mass of 0.0308 moles of P4

mass = number of moles * molar mass

molar mass of P4 = 4 * atomic mass of P = 4 * 31 g/mol = 124 g/mol

mass of P4 = 0.0308 moles * 124 g/mol = 3.8192g ≈ 3.82 g.

Answer: 3.82 grams of P4 will have the same number of molecules as 7.88 g of S8 (that is 0.0308 moles of molecules)
6 0
2 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
2 years ago
A 5.00 L sample of helium expands to 12.0 L at which point the
mina [271]

Answer:

1.73 atm

Explanation:

Given data:

Initial volume of helium = 5.00 L

Final volume of helium = 12.0 L

Final pressure = 0.720 atm

Initial pressure = ?

Solution:

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

P₁ × 5.00 L = 0.720 atm × 12.0 L

P₁ = 8.64 atm. L/5 L

P₁ = 1.73 atm

7 0
2 years ago
What is the mass percent of hydrogen in hexanal?
yKpoI14uk [10]
12.1% is the mass percent of hydrogen in hexanal
5 0
2 years ago
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If a collision between molecules is very gentle the molecules are
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If  collision  between   molecules  is  very gentle  the   molecules  are  more   likely  to   bound  without  reaction

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ollision   is  process  in  which  molecules  come  together   or  collide   with  one  another.Molecules  must  collide  with  sufficient   energy  so  that   chemical  bond can break.  In  addition  for  collision  to  occur  there  must  have  favorable  orientation.
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