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4 years ago

The percent yield for a chemical reaction can not be calculated until the reaction is completed

Chemistry

1 answer:

maw [93]4 years ago

5 0

Answer:

False

Explanation:

using the measurements of the amounts of chemicals, it can be calculated to some degree of accuracy depending mainly on the accuracy of the measurements and enviroments of the reaction

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1. 10.0 grams of water in the form of a solid has a temperature of 0.0 °C. This cube of

Ainat [17]

The initial 10.0 grams of water is a kind of intensive property.

What do you mean by intensive properties?

Intensive properties are those properties which do not depend upon the amount of the given substance. For example Temperature and pressure.

Generally, grams are used to represent concentrations. Therefore, always consider extensive. However, in the given question, the concentration is specified i.e. means fixed. So, the 10.0 grams of water become intensive in the given conditions.

Hence, the answer is an intensive property.

To learn more about the intensive property concept Check the given link

brainly.com/question/1496355

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4 0

1 year ago

Describe how electron movement is related to the bonding in methane, CH4.

lesya [120]

B) The atomic orbitals overlap and share electrons to form covalent bonds.

4 0

4 years ago

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Given the following balanced equation at 120°C: A(g) + B(g) ⇋ 2 C(g) + D(s)(a) At equilibrium a 4.0 liter container was found to

BlackZzzverrR [31]

Answer:

a) kc = 0,25

b) [A] = 0,41 M

c) [A] = 0,8 M

[B] =0,2 M

[C] = 0,2M

Explanation:

The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.

Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium.

If global reaction is:

A(g) + B(g) ⇋ 2 C(g) + D(s)

The kc = $\frac{[C]^2}{[A][B]}$

a) The concentrations of each compound are:

[A] = $\frac{1,60 moles}{4,0 L}$ = 0,4 M

[B] = $\frac{0,40 moles}{4,0 L}$ = 0,1 M

[C] = $\frac{0,40 moles}{4,0 L}$ = 0,1 M

kc = $\frac{[0,1]^2}{[0,4][0,1]}$ = 0,25

b) The addition of B and D in the same amount will, in equilibrium, produce these changes:

[A] = $\frac{1,60-x moles}{4,0 L}$

[B] = $\frac{0,60-x moles}{4,0 L}$

[C] = $\frac{0,60+2x moles}{4,0 L}$

0,25 = $\frac{[0,60+2x]^2}{[1,60-x][0,60-x]}$

You will obtain

3,75x² +2,95x +0,12 = 0

Solving

x =-0,74363479081119 → No physical sense

x =-0,043031875855476

Thus, concentration of A is:

$\frac{1,60-(-0,04 moles)}{4,0 L}$ = 0,41 M

c) When volume is suddenly halved concentrations will be the concentrations in equilibrium over 2L:

[A] = $\frac{1,60 moles}{2,0 L}$ = 0,8 M

[B] = $\frac{0,40 moles}{2,0 L}$ = 0,2 M

[C] = $\frac{0,40 moles}{2,0 L}$ = 0,2M

I hope it helps!

8 0

4 years ago

Read 2 more answers

George is writing a newspaper article about the fire at the local library. His introductory paragraph has an example of conserva

Lina20 [59]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

Write a balanced net ionic equation showing only the particles involved in the reaction and balancing the charges.Pb(s) AgNO3aq)

Alexus [3.1K]

Answer:

Pb(s) + 2Ag+(aq) → 2Ag(s) + Pb2+(aq)

Explanation:

Step 1: The unbalanced equation

Pb(s) + AgNO3aq) → Ag(s) + Pb(NO3)2(aq)

Step 2: Balancing the equation

Pb(s) + AgNO3aq) → Ag(s) + Pb(NO3)2(aq)

On the right side we have 2x NO3, on the left side we have 1x NO3.

To balance the amount of NO3 on both sides, we have to multiply AgNO3 (on the left side) by 2.

Pb(s) + 2AgNO3aq) → Ag(s) + Pb(NO3)2(aq)

On the left side we have 2x Ag, on the right side we have 1x Ag.

To balance the amount of Ag on both sides, we have to multiply Ag (on the right side) by 2. Now the equation is balanced.

Pb(s) + 2AgNO3(aq) → 2Ag(s) + Pb(NO3)2(aq)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will look like this

Pb(s) + 2Ag+(aq) → 2Ag(s) + Pb2+

8 0

4 years ago