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Radda [10]
3 years ago
11

In a laboratory experiment, john uses a message to seperate soil particles from water . Which technique separation which techniq

ue of separation is he using
Chemistry
1 answer:
eimsori [14]3 years ago
3 0
So visualize soil mixed with water. To separate it, you can use a sieve to filter the water.
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Chlorine is used to disinfect swimming pools. the accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chl
frez [133]

The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.

The volume of water is 2.29\times 10^{4} gal.

Since, 1 gal= 3785.41 mL

Thus, 2.29\times 10^{4} gal=2.29\times 10^{4}\times 3785.41 mL=8.66\times 10^{7}mL

Density of water is 1 g/mL thus, mass of water will be 8.66\times 10^{7}g.

Since, 1 grams of chlorine →10^{6} grams of water.

1 g of water →10^{-6} g of chlorine and,

8.66\times 10^{7}g of water →86.6 g of chlorine

Since, the solution is 9% chlorine by mass, the volume of solution will be:

V=\frac{100}{9}\times 86.6 mL=9.62\times 10^{2} mL

Thus, volume of chlorine solution is 9.62\times 10^{2} mL.

6 0
3 years ago
Which compound reacts with an acid to produce water and a salt
Hatshy [7]
I'm not exactly sure which one but I do know that an acid and a base react in a aqueous solution to form water, so i would probably eliminate the ones that aren't aqueous solutions.
3 0
3 years ago
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Acids are described as corrosive because they a. turn litmus blue. b. taste bitter. c. “eat away” at other materials. d. feel sl
Masteriza [31]

Answer:

C

Explanation:

acids are corrosive since they tend to destroy every single thing they do get a big example is the acidic rain which tends to corrode iron sheet thus making them to appear worn out and full of rust

4 0
3 years ago
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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
Balance this equation. Not all spots have to have a number.
Lisa [10]

2Na+2H2O----2NaOH+H2

3 0
3 years ago
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