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3 years ago

9

Create a chemical equation for "Bubbling chlorine gas through a solution of sodium bromide gives elemental bromine and a solutio

n of sodium chloride" and balance it.

1 answer:

Alexxandr [17]3 years ago

5 0

Answer:

$Cl_{2} +2NaBr ---> 2NaCl + Br_{2}$

Explanation:

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Calculate the volume of liquid in the tank sketched below. Give your answer in liters, and round to the nearest 0.1 L.

tensa zangetsu [6.8K]

Answer:

The answer is 18.9

Explanation:

3 0

3 years ago

How many particles are in a 151 g sample of Li2O?

neonofarm [45]

Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

To convert this mass value to moles, we use;

moles = mass/molar mass

Molar mass of Li2O = 6.9(2) + 16

= 13.8 + 16

= 29.8g/mol

Mole = 151/29.8g

mole = 5.07moles

number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23

= 30.52 × 10^23

= 3.052 × 10^24 particles.

4 0

3 years ago

Magnesium has three naturally occurring isotopes with masses of 23.99 amu, 24.99 amu, and 25.98 amu and natural abundances of 78

mihalych1998 [28]

Answer:24.31

Explanation:Contribution made by isotope of mass 23.99= 23.99×78.99=1894.97

Contribution made by isotope of mass 24.99=24.99×10.00=249.9

Contribution made by isotope of mass 25.98=25.98×11.01=286.04

Total contribution=1894.97+249.9+286.04=2430.91

Average mass=2430.91÷100

=24.31

3 0

3 years ago

Nitrogen (N2) and hydrogen (H2) react to make NH3 according to the following

Schach [20]

The number of moles of NH3 that could be made would be 0.5 moles

<h3>Stoichiometric reactions</h3>

From the balanced equation of the reaction:

N2 (g) + 3 H2(g) ----> 2NH3 (g)

The mole ratio of N2 to H2 is 1:3

Thus, for 0.50 moles of N2, 1.5 moles of H2 should be present. But 0.75 moles of H2 was allowed to react. Meaning that H2 is limiting in this case.

Mole ratio of H2 and NH3 = 3:2

Thus for 0.75 moles H2, the mole of NH3 that would be produced will be:

2 x 0.75/3 = 0.5 moles

More on stoichiometric calculations can be found here: brainly.com/question/8062886

3 0

2 years ago

A methanol-water mixture is to be flash distilled at 1 atm. If the feed is 25 mole %methanol, what are the liquid and vapor comp

frozen [14]

Answer:

Explanation:

Given that:

The distillation is carried out at a pressure of 1 atm

The feed harbors 25% mole of methanol (z)

The total moles of feed is usually 100 moles

In the system, we have both methanol and water

Using the total mole balance for the distillation column.

Fz = Lx + Vy

where;

F = amount of feed

z = mole fraction of ethanol (in feed)

L = amount of liquid product out of the column

V = amount of vapor product out of the column

x = mole fraction of methanol out of the liquid

y = mole fraction of methanol out of the vapor.

SO;

(a)

If all the feed is vaporized, then the vapor will likely have the same composition as the feed.

(b)

If no vaporization of the feed takes place, then the bottoms moving out of the column contains the same composition as the feed.

(c)

If 1/3 of the feed is vaporized; then 2/3 of the feed is at the bottom.

The balance equation would be:

$Fz = (\dfrac{2}{3}F) x + (\dfrac{1}{3}F)y \\ \\ z = \dfrac{2}{3}x+\dfrac{1}{3}y$

Replacing z = 0.25; we have:

$0.25 = \dfrac{2}{3}x+\dfrac{1}{3}y$

0.75 = 2x + y

(d)

If 2/3 of the feed is vaporized;

Then:

$Fz = (\dfrac{1}{3}F) x + (\dfrac{2}{3}F)y \\ \\ z = \dfrac{1}{3}x+\dfrac{2}{3}y$

replacing z = 0.25

$0.25 = \dfrac{1}{3}x+\dfrac{2}{3}y$

0.75 = x + 2y

6 0

2 years ago