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Stels [109]
3 years ago
11

Ethylene glycol is the main component in automobile antifreeze. To monitor the temperature of an auto cooling system, you intend

to use a meter that reads from 0 to 100. You devise a new temperature scale based on the approximate melting and boiling points of a typical antifreeze solution (-45oC and 115oC). You wish these points to correspond to 0oA and 100oA, respectively. What is a temperature of 45oC in oA
Chemistry
1 answer:
Lunna [17]3 years ago
4 0

Answer:

X = 27 .

Explanation:

Centigrade scale : melting point O⁰C , boiling point 100⁰C

New scale          melting point -45⁰C   , boiling point 115⁰C

Formula relating the two scale

\frac{C- 0 }{100-0} =\frac{X-(-45)}{115-( -45) }

\frac{C }{100} =\frac{(X+45)}{(115 +45) }

\frac{C }{100} =\frac{(X+45)}{160}

Given C = 45

X = ?

\frac{45 }{100} =\frac{(X+45)}{160}

X + 45 = 72

X = 27 .

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Find the volume of 0.130M sulfuric acid necessary to react completely with 65.9g sodium hydroxide
xxTIMURxx [149]

The volume of the 0.130 M sulfuric acid, H₂SO₄ required to react completely with 65.9 g sodium hydroxide, NaOH is 6.34 L

  • We'll begin by calculating the number of mole of in 65.9 g sodium hydroxide, NaOH. This can be obtained as follow:

Mass of NaOH = 65.9 g

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

<h3>Mole of NaOH =?</h3>

Mole = mass / molar mass

Mole of NaOH = 65.9 / 40

<h3>Mole of NaOH = 1.6475 mole</h3>

  • Next, we shall determine the number of mole of H₂SO₄ needed to react with 1.6475 mole of NaOH. This can be obtained as follow:

2NaOH + H₂SO₄ —> Na₂SO₄ + 2H₂O

From the balanced equation above,

2 moles of NaOH reacted with 1 mole of H₂SO₄.

Therefore,

1.6475 mole of NaOH will react with = \frac{1.6475}{2} \\\\ = 0.82375 mole of H₂SO₄.

  • Finally, we shall determine the volume of 0.130 M sulfuric acid, H₂SO₄ required for the reaction.

Molarity of H₂SO₄ = 0.130 M

Mole of H₂SO₄ = 0.82375 mole

<h3>Volume of H₂SO₄ =? </h3>

Volume = mole / Molarity

Volume of H₂SO₄ = 0.82375 / 0.130

<h3>Volume of H₂SO₄ = 6.34 L </h3>

Therefore, the volume of the 0.130 M sulfuric acid, H₂SO₄ required for the reaction is 6.34 L

Learn more: brainly.com/question/7882345

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