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laiz [17]
1 year ago
7

What volume of a 0.550 M solution of potassium hydroxide (KOH) can be made with 19.9 g of potassium hydroxide?

Chemistry
1 answer:
Natalija [7]1 year ago
6 0

Answer:

0.645 L

Explanation:

To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.

(Step 1)

Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol

Molar Mass (KOH): 56.104 g/mol

19.9 grams KOH              1 mole
--------------------------  x  -----------------------  =  0.355 moles KOH
                                     56.014 grams

(Step 2)

Molarity = moles / volume                            <----- Molarity ratio

0.550 M = 0.355 moles / volume                 <----- Insert values

(0.550 M) x volume = 0.355 moles              <----- Multiply both sides by volume

volume = 0.645 L                                          <----- Divide both sides by 0.550

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state two methods that can be used to prepare chlorine from rock salt.write an equation in each case​
Levart [38]

Answer:

electrolysis of brine

Explanation:

Rock salt deposits are usually mined; occasionally water is pumped down, and brine which contain 25 percent of sodium chloride is found

so d brine is electrolyzed to produce chlorine

7 0
2 years ago
What is the ph of a 7.5*10^-3 m H+ solution
natulia [17]
The pH+ of solution is 3
6 0
3 years ago
Read 2 more answers
A 0.100 M oxalic acid, HO2CCO2H, solution is titrated with 0.100 M KOH. Calculate the pH when 25.00 mL of oxalic acid solution i
____ [38]

Answer:

pH = 4.09

Explanation:

molarity of oxalic acid in the solution

         = 0.1 x 25 / (25 + 35)

         = 0.0417 M

molarity of NaOH in the solution

          = 0.1 x 35 / (25 +35)

         = 0.0583 M

H2C2O4 + NaOH -------------------> NaHC2O4 + H2O

0.0417          0.0583                            0                0

0                      0.0166                         0.0417

now second acid -base titration  

NaHC2O4 + NaOH -------------------> Na2C2O4 + H2O

0.0417             0.0166                              0                0

0.0251                 0                                  0.0166         ---

now

pH = pKa2 + log [Na2C2O4 / NaHC2O4]

pH = 4.27 + log (0.0166 / 0.0251)

pH = 4.09

3 0
2 years ago
Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate
Anvisha [2.4K]

Answer:

-3272     kJ/mol

Explanation:

Given and known facts

Mass of Benzene = 0.187 grams

Mass of water = 250 grams

Standard heat capacity of water = 4.18 J/g∙°C

Change in temperature ΔT = 7.48°C

Heat

=250 * 4.18 * 7.48\\=7816.6 \\=7.82

Heat released by benzine is - 7.82 kJ

Now, we know that

0.187 grams of benzene release = -7.82  kJ heat

So, 1 g benzine releases

\frac{ -7.82 }{0.187}\\= -41.8

kJ/g

0.187 * \frac{1}{78.108}=0.00239 mol C6H6

Heat released

= \frac{-7.82}{ 0.00239}

=-3272     kJ/mol

4 0
3 years ago
Grace wanted to find out the best conditions for growing lettuce plants.
Mariana [72]

Explanation:

the investigation lasts for 7 days !

hope this helps you.

8 0
2 years ago
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