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kotykmax [81]
3 years ago
11

If a gas has a volume of 3.67 L and a pressure of 790 mm Hg, what will the pressure be if the volume is compressed to 2.12 L? Wh

at is the pressure in atmospheres (atm)?
Chemistry
2 answers:
oksian1 [2.3K]3 years ago
7 0

If each gas sample has the same temperature and pressure, which has the greatest volume? Since hydrogen gas has the lowest molar mass of the set, 1 g will have the greatest number of moles and therefore the greatest volume. What is the Ideal Gas Law?

mrs_skeptik [129]3 years ago
3 0

Answer:

0.600

Explanation:

2.120/3.670 = 0.578 so 790*0.578 = 456.349 when converted to atm that is 456.349/760 = V2

0.600

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They are both water

Explanation:

One is cleaner than the other.

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There are moles of carbon present in 100 g of a
rodikova [14]

Answer: 3.33

Explanation:

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Katherine was given a tiny lamb by her uncle. The lamb was a runt and had to be fed by hand. Every 3 hours
lorasvet [3.4K]

Answer:

2.32 liters

Explanation:

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4 0
4 years ago
What volume (ml) of fluorine gas is required to react with 1. 28 g of calcium bromide to form calcium fluoride and bromine gas a
Alexandra [31]

144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

<h3>What is Ideal Gas Law ? </h3>

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Presure

V = Volume in liters

n = number of moles of gas

R = Ideal gas constant

T = temperature in Kelvin

Here,

P = 1 atm  [At STP]

R = 0.0821 atm.L/mol.K

T = 273 K  [At STP]

Now first find the number of moles

F₂  +  CaBr₂  →  CaF₂  +  Br₂

Here 1 mole of F₂ reacts with 1 mole of CaBr₂.

So,  199.89 g CaBr₂ reacts with  = 1 mole of F₂

1.28 g of CaBr₂ will react with = n mole of F₂

n = \frac{1.28\g \times 1\ \text{mole}}{199.89\ g}

n = 0.0064 mole

Now put the value in above equation we get

PV = nRT

1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K

V = 0.1434 L

V ≈ 144 mL

Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

Learn more about the Ideal Gas here: brainly.com/question/20348074

#SPJ4

4 0
2 years ago
The end point of a titration was reached after 22 ml of 0.050 m disodium edta titrant was dispensed into a solution containing t
ale4655 [162]
The moles  of disodium edta  used is  calculated  using  the  below  formula


moles =molarity  x  volume in  liters

molarity=0.050m
volume  in  liters = 22/1000=0.022 L

moles  is  therefore=  0.022 x0.050  =1.1  x10^-3  moles  of  disodium  edta


4 0
3 years ago
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