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2 years ago

There are two cups of water. The cups of water must have the same average Kinetic energy if they have the same ___

Chemistry

1 answer:

Oxana [17]2 years ago

6 0

Answer:

Temperature and Heat

Explanation:

Proportional to the average kinetic energy of the molecules in a substance.

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What is the pH of a 0.1 M phosphate buffer (pKa = 6.86) that contains equal amounts of acid and conjugate base?

Anvisha [2.4K]

Answer : The pH of a 0.1 M phosphate buffer is, 6.86

Explanation : Given,

$pK_a=6.86$

Concentration of acid = 0.1 M

Concentration of conjugate base (salt) = 0.1 M

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$pH=6.86+\log (\frac{0.1}{0.1})$

$pH=6.86$

Therefore, the pH of a 0.1 M phosphate buffer is, 6.86

4 0

2 years ago

solution of 0.160 MNaOH is used to titrate 21.0 mL of a solution of H2SO4:H2SO4(aq)+2NaOH(aq)→2H2O(l)+Na2SO4(aq)If 38.8 mL of th

stealth61 [152]

Answer:

0.0432 M H2SO4

Explanation:

First, we want to find the moles of MNaOH used. We know that Molarity x Liters = moles. 0.160M x 0.0210L = 0.00336 moles MNaOH

to find the moles of H2SO4, we can use a mol ratio.

0.00336mol MNaOH x (1Mol H2SO4 /2mol MNaOH)

= 0. 00168 mol H2SO4

I found the mol ratio by looking at the coefficients in front of the molecules I knew(MNaOH) and the molecule I needed to find(H2SO4)

then, to find Molarity, we do mol/Liters

0.00168 mol/ 0.0388L =. 0.0432 M H2SO4

You can convert mL to L by dividing by 1000

the significant figures of this problem is 3, so my final answer will also have 3 sig figs.

3 0

3 years ago

C(s) + S8(s) → CS2(l)

Vanyuwa [196]

The balanced chemical reaction is written as:

<span>4C(s) + S8(s) → 4CS2(l)

We are given the amount of carbon and sulfur to be used in the reaction. We need to determine first the limiting reactant to be able to solve this correctly.

</span>7.70 g C ( 1 mol / 12.01 g) =0.64 mol C
19.7 g S8 ( 1 mol / 256.48 g) = 0.08 mol S8

The limiting reactant would be S8. We use this amount to calculate.

0.08 mol S8 ( 4 mol CS2 / 1 mol S8 ) ( 256.48 g / 1 mol ) = 78.8 g CS2

4 0

2 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

6 0

2 years ago

How do you solve the level of a liquid in a graduated cylinder

Nezavi [6.7K]

Don’t really understand what you’re asking but, if you’re asking how to read a graduated cylinder:

Look at the graduated cylinder at eye level, find the meniscus, whatever the meniscus is at is your answer.

8 0

2 years ago