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Mandarinka [93]
2 years ago
5

Can anyone help with number 8.

Chemistry
1 answer:
Stels [109]2 years ago
7 0
A tornado! I think or it could be rain
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What is the new concentration of a solution of CaSO3 if 10.0 mL of a 2.0 M CaSO3 solution is diluted to 100 ml?
kifflom [539]

Answer: The new concentration of a solution of CaSO_{3} is 0.2 M 10.0 mL of a 2.0 M CaSO_{3} solution is diluted to 100 mL.

Explanation:

Given: V_{1} = 10.0 mL,      M_{1} = 2.0 M

V_{2} = 100 mL,           M_{2} = ?

Formula used to calculate the new concentration is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\10.0 mL \times 2.0 M = M_{2} \times 100 mL\\M_{2} = 0.2 M

Thus, we can conclude that the new concentration of a solution of CaSO_{3} is 0.2 M 10.0 mL of a 2.0 M CaSO_{3} solution is diluted to 100 mL.

5 0
2 years ago
What is analyse agro alemntary
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Answer:

subscribe other yt chanel

Explanation:

it will help you tu became big

4 0
2 years ago
What percentage of the earth is water?
OleMash [197]

Answer:

approximately 71%

4 0
3 years ago
Read 2 more answers
Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or nega
algol [13]

Answer: a) Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s):  negative

b)  CaCO_3(s)\rightarrow CaO(s)+CO_2(g) : positive

c) 2NH_3(g)\rightarrow N_2(g)+3H_2(g): positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s) : negative

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g):  positive.

f) I_2(s)\rightarrow I_2(g) : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

\Delta S is positive when randomness increases and \Delta S is negative when randomness decreases.

a) Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)

As ions are moving to solid form , randomness decreases and thus sign of \Delta S is negative.

b) CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

c) 2NH_3(g)\rightarrow N_2(g)+3H_2(g)

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of \Delta S is positive.

d) P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)

As gas is changing to solid, randomness decreases and thus sign of \Delta S is negative.

e) C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of \Delta S is positive.

f) I_2(s)\rightarrow I_2(g)

As solid is changing to gas, randomness increases and thus sign of \Delta S is positive.

6 0
3 years ago
Describe each highlighted bond in terms of the overlap of atomic orbitals. (If the highlighted bond is not a pi bond, select the
Anestetic [448]

The image of the bonds are missing, so i have attached it.

Answer:

A) - Sigma bond

-Sp³ and Sp³

- None

B) - Sigma and pi bond

- Sp² of C and p of O

- p of C and P of O

Explanation:

A) For compound 1;

- the molecular orbital type is sigma bond due to the end-to-end overlapping.

- Atomic orbitals in the sigma bond will be Sp³ and Sp³

- Atomic orbitals in the pi bond would be nil because there is no pi bond.

B) For compound 2;

- the molecular orbital type is sigma and pi bond

-Atomic orbitals in the sigma bond would be Sp² of C and p of O

- The Atomic orbitals in the pi bond will be; p of C and p of O

6 0
2 years ago
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