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Anna71 [15]
3 years ago
13

What is five plus forty five

Chemistry
2 answers:
zhenek [66]3 years ago
8 0

Answer:

55. I hope this helps you!

ziro4ka [17]3 years ago
5 0

Answer:

5+ 45 = 50 is the answer....

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Phosgene, a poisonous gas, when heated will decompose into carbon monoxide and chlorine in a reversible reaction: COCl2 (g) <
Advocard [28]

Answer:

8.08 × 10⁻⁴

Explanation:

Let's consider the following reaction.

COCl₂(g) ⇄ CO (g) + Cl₂(g)

The initial concentration of phosgene is:

M = 2.00 mol / 1.00 L = 2.00 M

We can find the final concentrations using an ICE chart.

     COCl₂(g) ⇄ CO (g) + Cl₂(g)

I       2.00            0            0

C        -x             +x           +x

E    2.00 -x          x             x

The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.

The concentrations at equilibrium are:

[COCl₂] = 2.00 -x = 1.96 M

[CO] = [Cl₂] = 0.0398 M

The equilibrium constant (Keq) is:

Keq = [CO].[Cl₂]/[COCl₂]

Keq = (0.0398)²/1.96

Keq = 8.08 × 10⁻⁴

4 0
3 years ago
Covalent solutes are considered non-electrolytes. What does this mean for the conductivity of the solution? A) Non-electrolytes
iogann1982 [59]

Electrolytes are those which dissociates in solution and produces ions.

Ions can carry current,so Electrolytes conduct electiricity.

And non electrolytes are those which do not dissociate in solution and doesnt produce ions.

Since non electrolytes do not produce ions they cannot conduct electricity.

<u>Hence the right option is:</u>

B) Non-electrolytes dissolve and do not dissociate in water providing no charged ions to conduct electricity.  

5 0
3 years ago
Read 2 more answers
If a neutral atom has 10 proton how many electrons would it have?
olchik [2.2K]
It should have 10 electrons
7 0
3 years ago
Read 2 more answers
A compound is 44.82% Potassium, 18.39% Sulfur, and 36.79% Oxygen. Write the empirical formula and name the compound.
sleet_krkn [62]

Answer:

Empirical formula = K₂SO₄

The name of the compound is Potassium tetraoxosulfate (vi)

Explanation:

Percentage composition of the elements:

Potassium = 44.82%

Sulfur = 18.39%

Oxygen = 36.79%

Mole ratio of the elements:

Potassium = 44.82/39 =1.15

Sulfur = 18.39/32 = 0.57

Oxygen = 36.79/16 = 2.29

Dividing by the smallest ratio in order to obtain the simplest ratio

Potassium = 1.15/0.57 =2

Sulfur = 0.57/0.57 = 1

Oxygen = 2.29/0.57 = 4

Therefore, the mole ratio of Potassium : Sulfur : Oxygen = 2:1:4

Empirical formula = K₂SO₄

The name of the compound is Potassium tetraoxosulfate (vi)

5 0
2 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
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