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3 years ago

The boiling point of acetone is much greater than the boiling point of 2-methylpropane.

Chemistry

1 answer:

Andreyy893 years ago

3 0

Answer:

True.

Explanation:

The boiling point of acetone is 56°C while boiling point of 2-methyl propane is -11°C. Based on these values, <em>boiling point of acetone is much greater than the boiling point of 2-methylpropane. </em>

That could be because acetone has an electronegative atom (Oxygen) doing high intermolecular forces in this molecule while 2-methylpropane is an apolar molecule.

I hope it helps!

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he decomposition of acetaldehyde, CH3CHO, was determined to be a second order reaction with a rate constant of 0.0771 M-1 s-1. I

nataly862011 [7]

Answer:

The concentration is [-1 + sqrt(1+0.11t)]/0.1542 M

Explanation:

Let the concentration of CH3CHO after selected reaction times be y

Rate = Ky^2 = change in concentration of CH3CHO/time

K = 0.0771 M^-1 s^-1

Change in concentration of CH3CHO = 0.358 - y

0.0771y^2 = 0.358-y/t

0.0771ty^2 = 0.358 - y

0.0771ty^2 + y - 0.358 = 0

The value of y must be positive and is obtained in terms of t using the quadratic formula

y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M

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3 years ago

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mezya [45]

Answer:

See explanation

Explanation:

The law of conservation of mass states that mass can neither be created nor destroyed. This implies that in a chemical reaction, we can only have the same number of atoms of each element on both sides of the reaction equation.

If we write 4Fe2S3 it means that we have;

4 * 2 = 8 atoms of Fe

4 * 3 = 12 atoms of S

8 + 12 = 20 atoms in all

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3 years ago

What is electrons configuration for arsenic ?

Anna35 [415]

Answer:

[Ar] 3d¹⁰ 4s² 4p³

Explanation:

so...... ur welcome

7 0

3 years ago

6.00g of gold was heated from 20.0c to 22.0c. How much heat was applied to the gold

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6 x .129 x 2= 1.55J
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2 years ago

7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$

Cross multiply

$0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4$

Divide both side by 0.505

$V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL$

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

$1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}$

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

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3 years ago