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kvv77 [185]
3 years ago
11

Answer this question pleas

Chemistry
2 answers:
Korvikt [17]3 years ago
4 0

Answer:nope

Explanation:

Just no

Aliun [14]3 years ago
4 0
Is really easy just do it by yourself and you’ll know great advice right lol
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harina [27]

Each orbital must contain a single electron before any orbital contains two electrons.

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Drag each label to the correct location on the image identify the parts of the energy diagram
diamong [38]
If you would’ve attached a picture I’m sure it would’ve been a lot easier.
4 0
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K2CrO4 is added to a solution containing lead (II) and barium ions. If a precipitate is formed, what is it?
Step2247 [10]

Answer:

Lead (II) Chromate and Barium Chromate

Explanation:

5 0
2 years ago
If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?
Alex73 [517]

Answer:

0.189 g.

Explanation:

  • This problem is an application on <em>Henry's law.</em>
  • Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
  • Solubility of the gas ∝ partial pressure
  • If we have different solubility at different pressures, we can express Henry's law as:

<em>S₁/P₁ = S₂/P₂,</em>

S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ??? g/L and P₂ = 5.73 atm

  • So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>

<em></em>

8 0
3 years ago
28) Consider a 21.0 mL sample of pure lemon juice with a citric acid (H3C6H5O7) concentration of 0.30M. a. How many moles of cir
Damm [24]
<h3>#a. Answer:</h3>

0.0063 mole

<h3>Solution and explanation:</h3>

We are given 21.0 mL citric acid with a concentration of 0.30 M

Part a requires we calculate the number of moles of citric acid.

We need to know how to calculate the concentration of a solution;

Concentration or molarity = Number of moles ÷ Volume of the solution

Thus;

Number of moles = Concentration × Volume

Hence;

Moles = 0.30 M × 0.021 L

         = 0.0063 mole

<h3>#b. Answer</h3>

1.21 g citric acid

<h3>Solution</h3>

Part B

We are required to calculate the mass of citric acid in the sample

Number of moles of a compound is calculated by dividing its mass by its molar mass.

Molar mass of Citric acid = 192.124 g/mol

Moles of citric acid = 0.0063 mole

But; Mass = Number of moles × Molar mass

Mass of citric acid = 0.0063 mol × 192.124 g/mol

                             = 1.21 g citric acid

<h3>#c. Answer</h3>

4.167 mL

<h3>Solution:</h3>

Part C

We are required to determine the initial volume before dilution;

We have;

Initial concentration (M1) = 0.30 M

Final volume (V2) = 250 mL or 0.25 L

Final concentration (M2) = 0.0050 M

Using the dilution formula we can get the initial volume;

Therefore, since; M1V1 =M2V2

V1 = M2V2÷M1

   = (0.0050 × 0.25)÷ 0.30

   = 0.004167 L or

   = 4.167 mL

Therefore, the initial volume of the solution is 4.167 mL

8 0
3 years ago
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