Mr Reams uses 10 cups of paint to fill the small dishes, therefore, the amount of paints left in the bowl is 2 cups.
<h3>How to convert ounces to cups?</h3>
He mixes 12 cups of paints in a large bowl.
He uses the paint to fill 40 small dishes with 2 fluid ounces of paint each.
1 small dish = 2 fluid ounces
40 small dishes = 80 fluid ounces
Therefore,
1 fluid 0unces = 0.125 cups
80 fluid ounces = ?
cross multiply
amount of paint filled = 80 × 0.125 = 10 cups
Therefore, the amount of cups pf paint Mr Reams have left in the bowl is 12 cups - 10 cups = 2 cups.
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Answer:B Gates of Paradise
Explanation: I got it right and cuz I said soooo
The calculated pH of the solutions are given below:
- The pH of the solution after 14.0 ml of base is added to it is calculated as 1.45.
- The pH of the solution after 19.8 ml of base is added to it is calculated as 3.0
- The pH of the solution after 20.0 ml of base is added to it is calculated as 7.
<h3>What is pH?</h3>
This is the level of acidity or alkalinity of an aqueous solution. The pH of a substances tells if its an acid, base or neutral.
HBr and NaOH while in water would dissociate and they would become H+ and OH- respectively.
Mol of HBr = Mol * Vol
= 0.200 * 20mL
= 4 mmol
A. Moles of NaOH added
= 0.2 X 14
= 2.8 mmol
Moles of H+ that did not react
= 4 - 2.8
= 1.2 mmol
1.2/1000 = 0.0012 moles
Total volume = 20 + 14
= 34 mL to litres
= 0.034 L
Molarity of H+ = 0.0012 / 0.034L
= 0.035 M
-log[0.035] = 1.45
The pH of the solution is 1.45
B. NaOH added = 19.8 * 0.2 =
3.96 mmoles
The unreacted solution
= 4.0 - 3.96
= 0.04 mmol
0.04/1000 = 0.00004 moles
Total volume = 20 + 19.8
= 39.8mL
Converted to litres = 0.0398L
Molarity = 0.00004 / 0.0398L
= 0.001
-log(0.001) = 3.0
The ph Is therefore 3.0
C. Moles of NaOH added = 0.2*20mL = 4mmol
All the H+ are going to be consumed here. This would result into a neutral solution pH = 7
<u>Complete question:</u>
20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the PH of the solution after the following volumes of base have been added.
A. 14.0 mL
B. 19.8 mL
C. 20.0 mL
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