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AURORKA [14]
3 years ago
10

A balloon has a volume of 10,500 liters and the tempature is 15 C . If the temp tire were -25C what would the volume of the ball

oon be?
A-12,194L
B- 9,042L
Chemistry
1 answer:
Harlamova29_29 [7]3 years ago
5 0
Given an initial volume of 10,500 liters and initial temperature of 15 C, final temperature of -25 C, and assuming that the gas inside the balloon is an ideal gas:

Ideal Gas Law:

nRT = PV

assume that the number of moles and pressure are constant, the equation becomes

V/T = constant

V1/T1 = V2/T2

Substituting given conditions, solve for V2:

10,500L/ (15+273) = V2/ (-25 + 273)

V2 = 9041.67 L
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Need help with this ASAP<br><br> Thanks!
olga_2 [115]

Answer:

Decomposers (either Secondary Consumer or Tertiary Consumer)

Explanation:

Decomposers eat dead materials and break them down into chemical parts. ... They keep the ecosystem free of the bodies of dead animals or carrion. They break down the organic material and recycle it into the ecosystem as nutrients. Vultures, Blowflies, hyenas, crabs, lobsters and eels are examples of scavengers.

6 0
3 years ago
This is very helpful, but what is the best one?
Mice21 [21]
Can you tell us what you mean and post the questions ?
8 0
4 years ago
Select the correct answer. What is the percent composition of silicon in silicon carbide (SiC)? A. 28% B. 50% C. 70% D. 142%
Nadya [2.5K]

Answer:

C. 70%

Explanation:

Atomic Mass of the silicon = 28 g.

Atomic mass of the Carbon = 12 g.

Total mass of the Silicon Carbide = 28 + 12

= 40 g.

Now, Using the formula.

% Composition = Mass of the silicon/Total mass of the compound × 100 %

= 28/40 × 100 %

= 70 %

Hence, % composition of the silicon in SiC is 70%

7 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
3 years ago
Help please
Lana71 [14]

Answer:

Yes

Explanation:

Based on the graph, nuclear energy is one of the least contributors of CO2 emissions.

You would support this source of energy .

8 0
3 years ago
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