You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
Answer:
49.95 g of HCl
Explanation:
Let's formulate the chemical equation involved in the process:
Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O
This means that we need 1 mole of Calcium hydroxide to neutralize 2 moles of hydrochloric acid. From this, we calculate the quantity of HCl moles that would be neutralized by 0.685 moles of Ca(OH)2
1 mole Ca(OH)2 ---- 2 moles HCl
0.685 moles Ca(OH)2 ---- x = 1.37 moles HCl
Now that we know the quantity of HCl moles that would react, let's calculate the quantity of grams this moles represent:
1 mole of HCl ---- 36.46094 g
1.37 moles ------ x = 49.95 g of HCl
<span>31.9 grams butane needed to produce 96.7 grams CO2
</span>
Answer:
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