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Fed [463]
3 years ago
9

A gas sample is at 25°C and 1.0 atmosphere. Which changes in temperature and pressure will cause this sample to behave more like

an ideal gas?
(1) decreased temperature and increased pressure
(2) decreased temperature and decreased pressure
(3) increased temperature and increased pressure
(4) increased temperature and decreased pressure
Chemistry
1 answer:
-BARSIC- [3]3 years ago
3 0
What's the answer. tell me please please please please please please please please please please please please 
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HELP!!!
vredina [299]

Answer:

the answer is A

Explanation:

I dont really have an explanation but hope it helps

5 0
2 years ago
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Please help do not get thanks
slega [8]
5. A
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6 0
3 years ago
What happens to solubility when the solute is crushed before adding to water
marusya05 [52]
It will dissolve faster as the surface area is more so more particles collide ---> faster dissolving
6 0
3 years ago
Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, y
denis23 [38]

Answer:

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

Explanation:

Given that:

the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of CHCl_3 = 1.492 g/mL

The density of CHBr_3 = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of  CHCl_3 = P ml

and the volume of CHBr_3 = Q ml

the sum of their volumes should be equal to the total volume of the mixture

P \ ml + Q \ ml = 20 ml  ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

\mathtt{(Density \ of  \ CHCl_3 \times Vol. \ of  \ CHCl_3 ) + (Density  \  of \  CHBr_3  \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density  \ of \ mixture \times volume \ of \ the \ mixture)}

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL  ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g =  36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

6 0
3 years ago
9A. A sample of hydrogen at 1.56 atm had it's pressure decreased to 0.73
densk [106]

Answer:

351.43mL

Explanation:

To calculate the original volume of hydrogen gas in this question, the Boyle's law equation will be used. Boyle's law equation is:

P1V1 = P2V2

Where; P1 = initial pressure

V1 = initial volume

P2 = final pressure

V2 = final volume

According to this question, the P1= 1.56atm, V1 = ?, P2 = 0.73atm, V2 = 751mL

Hence;

P1V1 = P2V2

1.56 × V1 = 0.73 × 751

1.56 V1 = 548.23

V1 = 548.23/1.56

V1 = 351.43mL

Therefore, the original volume of hydrogen gas is 351.43 mL.

4 0
3 years ago
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