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valentinak56 [21]
2 years ago
11

A sample of gas is placed into an enclosed cylinder and fitted with a movable piston. Calculate the work (in joules) done by the

gas if it expands from 5.33 L to 11.05 L against a pressure of 1.50 atm.
Chemistry
1 answer:
Nuetrik [128]2 years ago
6 0

Explanation:

here is the answer. Feel free to ask for more chem help

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What is the solution to the problem expressed to the correct number of significant figures? 12.0/7.11=?
xxMikexx [17]

Answer:

1.69.

Explanation:

  • The solution = 12.0 / 7.11 = 1.687 = 1.69.
  • The rule of significant figures for division states that: the results are reported to the fewest significant  figures.
  • 12.0 contains 3 significant figures.
  • 7.11 contains 3 significant figures.

So, the solution should contain 3 significant figures.

  • Now, the issue id of rounding; In a series of calculations, carry the extra digits through to  the final result, then round.
  • If the digit to be removed is equal to or greater than 5, the preceding digit is  increased by 1.
  • The digit that should be removed is 7 that is larger than 5 so increase the preceding digit by 1.
3 0
3 years ago
Please help! BRAINLIEST to right answerrrr
zloy xaker [14]

Answer: from the hotter surface to the colder one

Explanation:

8 0
3 years ago
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The volume of a fixed amount of gas is doubled, and the absolute temperature is doubled. According to the ideal gas law, how has
neonofarm [45]

Answer:

Option A. It has stayed the same.

Explanation:

To answer the question given above, we assumed:

Initial volume (V₁) = V

Initial temperature (T₁) = T

Initial pressure (P₁) = P

From the question given above, the following data were:

Final volume (V₂) = 2V

Final temperature (T₂) = 2T

Final pressure (P₂) =?

The final pressure of the gas can be obtained as follow:

P₁V₁/T₁ = P₂V₂/T₂

PV/T = P₂ × 2V / 2T

Cross multiply

P₂ × 2V × T = PV × 2T

Divide both side by 2V × T

P₂ = PV × 2T / 2V × T

P₂ = P

Thus, the final pressure is the same as the initial pressure.

Option A gives the correct answer to the question.

3 0
3 years ago
The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A
s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $ m^3 /d $

                                    $= 80 \times 10^5 \ I/d $

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L $

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d $

If 60 mg of alum $ [Al_2(SO_4)_3.14 H_2O] $ required for one litre of the water treatment.

So Alum required for  $ 80 \times 10^5 \ I/d $

$= 80 \times 15^5 \ I/d  \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d $

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $ = 60 \times 0.234 \text{ of alum ppt. per litre} $

      $= 14.04 $ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

                                        = 112.32 kg/d ppt of alum

Daily total solid load is  $= 144 \ kg/d + 112.32 \ kg/d$

                                       = 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234 $

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

 The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-} $

After addition, the aluminium hydroxide pH of water will increase due to increase in $ OH^- $ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm $

And the pH is reduced into the range of 5.9 to 6.4

6 0
2 years ago
How many moles of N2 are need to fill a 35 L tank at standard temperature and pressure?
Tems11 [23]

1.56 moles of N2 are needed to fill a 35 L tank at standard temperature and pressure. Details about moles can be found below.

<h3>How to calculate number of moles?</h3>

The number of moles of a substance can be calculated using the following formula:

PV = nRT

Where;

  • P = pressure
  • V = volume
  • n = number of moles
  • R = gas law constant
  • T = temperature

At STP;

  • T = 273K
  • P = 1 atm
  • R = 0.0821 Latm/molK

1 × 35 = n × 0.0821 × 273

35 = 22.41n

n = 35/22.41

n = 1.56mol

Therefore, 1.56 moles of N2 are needed to fill a 35 L tank at standard temperature and pressure.

Learn more about number of moles at: brainly.com/question/14919968

#SPJ1

4 0
2 years ago
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