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Deffense [45]
3 years ago
10

How many milliliters of 2.0M Mg(OH)2 are required to neutralize 250.0 ml of 0.80M hydrochloric acid (HCl)?

Chemistry
1 answer:
Juliette [100K]3 years ago
4 0

Answer:

OK, This is my thinking, I hope this helps you out.

Let's do problem "a"

Step 1.  Write a balanced equation.

                                       2 HCl  +  Ba(OH)2  -------------->  BaCl2  + 2 HOH  

Step 2.  Underneath the HCl and Ba(OH)2 in the equation write what you are given and what you need to find out.

                                    2 HCl        +       Ba(OH)2  -------------->  BaCl2  + 2 HOH

                                 C = 0.130 mol/L    v = 56 mL

                                  V = ?                   C = 0.109 mol/L

Step 3.  They ALWAYS provide a way to calculate the number of moles of one of the substances.  In this case it is Ba(OH)2 because you have C and V

                         n = C X V            n = 0.109 x 56 = 6.104 millimoles

Step 4.  Use the equation to figure out how many moles of the unknown (HCl) you will need.

               the equation says you need TWICE as many moles as you have of Ba(OH)2.  So you will need 6.104 millimoles x 2 or 12.208 milimoles.

Step 5.  Now that yo have moles and concentration of HCl you can now calculate volume

                  C = n / V, so V = n / C

                                 12.208 millimoles / 0.130 mol/L = 93.9 mL

Use this method for all these kind of problems.  

In B) you will have to convert he g of NaOH into moles by 0.240 / 40.00 g/mol and then following trhe rest of the procedure.

C) is done the same way.  convert the g of Na2SO4 into moles first.

d) is almost the same as a) except this time when you find the moles of Ca(OH)2 you just have to convert the moles into mass by multiplying moles x Molar mass of Ca(OH)2.

Best of luck.

Explanation:

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Explanation:

If we assume that the pressure is constant and the number of moles does not change, we can say that the volume of the gas is modified in a directly ratio, to the Absolute Temperature.

Let's convert the values:

91°C + 273 = 364K

0.9°C + 273 = 273.9K

Volume decreases if the temperature is decreases

Volume increases if the T° increases

V₁ / T₁  = V₂ / T₂ → 1.63mL /364K = V₂ / 273.9K

V₂ = (1.63mL /364K) . 273.9K → 1.22 mL

5 0
3 years ago
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chubhunter [2.5K]

Answer:

ΔH of dissociation is 38,0 kJ/mol

Explanation:

The dissociation reaction of KBrO₃ is:

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This dissolution consume heat that is evidenced with the decrease in water temperature.

The heat consumed is:

q = CΔTm

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And m is mass of water (150,0 mL ≈ 150,0 g)

Replacing, heat consumed is:

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13,8 g of KBrO₃ are:

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<em></em>

I hope it helps!

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5 0
3 years ago
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enyata [817]

Answer:

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Explanation:

Given data:

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