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svetlana [45]
2 years ago
9

Please help i will give extra points

Chemistry
1 answer:
steposvetlana [31]2 years ago
4 0

Answer:

D) It is a solid

Explanation:

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Exactly 5000 mL of air at 223K is warmed and has a new volume of 8.36 liters. What is the new temperature?
34kurt

Answer:

The new temperature is 373 K

Explanation:

Step 1: Data given

Volume air = 5000 mL = 5.0 L

Temperature = 223K

New volume = 8.36 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒V1 = the initial volume = 5.0 L

⇒T1 = the initial temperature = 223 K

⇒V2 = the new volume = 8.36 L

⇒T2 = the new temperature

5.0/223 = 8.36 /T2

T2 = 373 K

The new temperature is 373 K

7 0
3 years ago
A buret was improperly read to 1 decimal place giving a reading of 27.2 mL. If the actual volume is 27.26 mL, what is the percen
Elenna [48]
Using the significant figure it would be 27.3
8 0
2 years ago
You have been shipwrecked on a deserted island with no running water/fresh water, you have your clothes and a plastic bag contai
artcher [175]

Answer:

i have no clue

Explanation:

6 0
2 years ago
Please help!
Dahasolnce [82]

Answer:

reproduction

Explanation:

reproduction, process by which organisms replicate themselves

6 0
2 years ago
A solution of barium nitrate has 61.2g of barium nitrate in 1 liter of solution. How many mg of barium are there in 7.5 quarts
Nuetrik [128]

Answer:

220.44g Ba²⁺ ions in solution

Explanation:

Given parameters:

Mass of barium nitrate = 61.2g

Volume of solution = 1 liter

Unkown:

Mass of barium in 7.5quarts of solution?

Solution

We must first convert quarts to its liter equivalence:

                    1 quarts = 0.95 liter

                   7.5 quarts = 0.95 x 7.5; 7.125liter

Now, let us find the mass of barium nitrate in a solution of 7.125liter:

        Given:

            1 liter of solution contains 61.2g of barium nitrate:

          7.125 liter will contain  7.125 x 61.2 = 436.05g of barium nitrate.

The formula of the compound is Ba(NO₃)₂:

  In solution we have  Ba²⁺ + NO₃⁻

                Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻

 

Number of moles of Ba(NO₃)₂ = \frac{mass of Ba(NO₃)₂}{Molar mass of Ba(NO₃)₂}

Molar mass of Ba(NO₃)₂ = 137 + 2[14 + 3(16)] = 271g/mol

Number of moles of Ba(NO₃)₂ = \frac{436.05}{271} = 1.609mole

      1 mole of Ba(NO₃)₂ will produce 1 mole of Ba²⁺ ions in solution

    therefore, 1.609mole of Ba(NO₃)₂ will also yield 1.609mole of Ba²⁺ ions in solution

Mass of Ba²⁺ ions = Number of moles of Ba²⁺ ions  x molar mass of Ba²⁺ ions

Mass of Ba²⁺ ions = 1.609 x 137 = 220.44g

3 0
3 years ago
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