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Inga [223]
3 years ago
13

PLEASEEEE HELLPPPP ASAP

Chemistry
1 answer:
sergij07 [2.7K]3 years ago
5 0
The answer is A

According to research I have done, pure solids and liquids are not included in the equilibrium constant expression. If the concentration of a reactant in aqueous solution is increased, the position of equilibrium will move in the direction which minimises the effect of this increase in concentration, by using the added component up, to decrese it's concentration again.
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How many grams of K2Cr207 are in 250. mL of 0.500M of K2Cr207:
goldenfox [79]

<u>Analysing the Question:</u>

We are given a 250 mL solution of 0.5M K₂Cr₂O₇

Which means that we have:

0.5 Mole in 1L of the solution

0.125 moles in 250 mL of the solution      <em>[dividing both the numbers by 4]</em>

<em />

<u>Mass of K₂Cr₂O₇ in the given solution:</u>

Molar mass of K₂Cr₂O₇(Potassium Dichromate) = 194 g/mol

<em>we know that we have 0.125 moles in the 250 mL solution provided</em>

Mass = Number of moles * Molar mass

Mass = 0.125 * 194

Mass = 36.75 grams

7 0
3 years ago
Muriatic acid, HCl, is often used to remove rust. A solution of muriatic acid, HCl, reacts with Fe2O3 deposits on industrial equ
Alina [70]

Answer:

3L

Explanation:

Step 1:

The balanced equation for the reaction.

Fe2O3(s) + 6HCl(aq) → 2FeCl3(aq) + 3H2O

Step 2 :

Determination of the masses of HCl and Fe2O3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Fe2O3 = 159.70g/mol

Molar mass of HCl = 36.46 g/mol

Mass of HCl from the balanced equation = 6 x 36.46 = 218.76g

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl

Step 3:

Determination of the mass of HCl needed to react with 439g of Fe2O3. This is illustrated below:

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl.

Therefore, 439g of Fe2O3 will react with = (439 x 218.76) /159.70 = 601.35g of HCl.

Step 4:

Conversion of 601.35g of HCl to mole. This is illustrated below:

Molar mass of HCl = 36.46 g/mol

Mass of HCl = 601.35g

Number of mole = Mass/Molar Mass

Number of mole of HCl = 601.35/36.46

Number of mole of HCl = 16.49 moles

Step 5:

Determination of the volume of the HCl that reacted.

This is illustrated below:

Mole of HCl = 16.49 moles

Molarity of HCl = 5.50 M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 16.49/5.5

Volume of HCl = 3L

Therefore the volume of HCl needed for the reaction is 3L

5 0
3 years ago
A certain metal M crystallizes in a lattice described by a body-centered cubic (bcc) unit cell. The lattice constant (a, the edg
Furkat [3]

Answer:

radius = 156 pm

Explanation:

The relation between radius and edge length of unit cell of BCC is

r=a\sqrt{3}/4

Given

a = 360 pm

Therefore

r = r = radius = 360\sqrt{3}/4= 155.88 pm

Or

156 pm

3 0
3 years ago
When 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4Cl, how many grams of AgCl are formed?
irina1246 [14]

Answer:

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

                 AgNO₃(aq)  +  NH₄Cl(aq)   ⇒   AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

moles = 0.0039

2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

moles = 0.00188

3.- Calculate the limiting reactant

The proportion of     AgNO₃(aq)  to  NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

                     1 mol of NH₄Cl  ---------------- 1 mol of AgCl

              0.00188 mol of NH₄Cl ------------- x

                     x = (0.00188 x 1) /1

                     x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

                         143.5 grams of AgCl -------------- 1 mol

                         x -------------------------------------------0.00188 moles of AgCl

                          x = (0.00188 x 143.5) / 1

                          x = 0.269 grams of AgCl

8 0
3 years ago
What quantity is represented by the metric system prefix deci-?
mario62 [17]
10. For example, decagon and decade.
3 0
3 years ago
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