Exemplificati solutii in care solventul este gaz

What would happen if a Continental Tropical Front meets a Maritime Polar front?

rewona [7]

Answer:

As soon as a front passes a particular location, the weather could change from a steamy maritime tropical air mass to a dry.

Explanation:

hope this helps

7 0

3 years ago

Read 2 more answers

The pressure of a 250 mL sample of gas is 105 kPa. What would be the pressure if the volume were increased to 375 ml?

Arte-miy333 [17]

Answer:

<h2>The answer is 70 kPa</h2>

Explanation:

In order to find the pressure if the volume were increased to 375 ml we use Boyle's law

That's

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the final pressure

$P_2 = \frac{P_1V_1}{V_2} \\$

From the question

P1 = 105 kPa = 105 , 000 Pa

V1 = 250 mL

V2 = 375 mL

So we have

$P_2 = \frac{105000 \times 250}{375} = \frac{26250000}{375} \\ = 70000$

We have the final answer as

Hope this helps you

8 0

4 years ago

Determine the number of unpaired electrons expected for [Fe(NO2)6]3−and for [FeF6]3− in terms of crystal field theory.

arsen [322]

Answer:

A. One unpaired electron

B. 5 unpaired electrons

Explanation:

In A ,Fe is in +3 oxidation state and Electronic configuration- [Ar]3d5

And NO2 is a strong field ligand hence it causes pairing in t2g orbitals and results one unpaired electron in dZX orbital.

In B, also Fe is in +3 oxidation state but F is weak field ligand hence causes no pairing of Electrons hence it results 5 unpaired electrons with electronic configuration t2g^3 eg^2

7 0

3 years ago

Consider the reaction below. Mg(s) + 2HCl(aq) mc019-1.jpg H2(g) + MgCl2(aq) What is the most likely effect of an increase in pre

bonufazy [111]

1) Reaction:

Mg(s) + 2HCl (aq) ----> H2(g) + MgCl2(aq)

2) Analysis

An increase in pressure affects directly the rate of reaction involving reactiong gases. Changing the pressure where there are only solids or liquids does not affect the rate of reaction.

This reaction is not an equilibrium, the reaction is only forward. So, the reacting components, Mg(s) and HCl(aq) are a solid and a liquid.

Therefore, the reaction is not affected by the change in pressure.

Answer: the reaction is not affected at all.

4 0

3 years ago

Read 2 more answers

Balance the following redox equations by the half-reaction method:

frozen [14]

Answer:

a) Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

b) 2Bi(OH)3 + 3SnO2^2- → 2Bi + 3SnO3^2- + 3H2O

c) Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

d) 2ClO3- + 2Cl- + 4H+ → Cl2 + 2H2O +2ClO2

e) 5 BiO3^- + 14 H+ + 2Mn^2+ → 5Bi^3+ + 7H2O + 2MnO4^-

Explanation:

<em>(a) Mn2 + H2O2 → MnO2 + H2O (in basic solution)</em>

Step 1: The half reactions

Oxidation: Mn2+ + 4OH- → MnO2 + 2H2O + 2e-

Reduction: H2O2 + 2e- + 2H2O → 2H2O + 2OH-

Step 2: Sum of both half reactions

Mn2+ + 4OH- + H2O2 → MnO2 + 2H2O + 2OH-

Step 3: the netto reaction

Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

<em>(b) Bi(OH)3 + SnO2^2- → SnO3^2- + Bi (in basic solution)</em>

Step 1: The half reactions

Reduction: Bi(OH)3 + 3e- → Bi

Oxidation : Sno2^2- → SnO3^2- +2e-

Step 2: Balance the half reactions

2* (Bi(OH)3 + 3e- → Bi + 3OH-)

3* (Sno2^2- +2OH- → SnO3^2- +2e- + H2O)

Step 3: The netto reaction

2Bi(OH)3 + 3SnO2^2- → 2Bi + 3SnO3^2- + 3H2O

<em>(c) Cr2O7^2- + C2O4^2- → Cr^3+ + CO2 (in acidic solution)</em>

Step 1: The half reactions

Reduction: Cr2O7^2- + 6e- → 2Cr+

Oxidation : C2O4^2- → 2CO2 + 2e-

Step 2: Balance the half reactions

Cr2O7^2- + 6e- +14H+ → 2Cr+ +7H2O

3*(C2O4^2- → 2CO2 + 2e-)

Step 3: The netto reaction

Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

<em>(d) ClO3^- + Cl^− </em>→<em> Cl^2 + ClO^2 (in acidic solution)</em>

Step 1: The half reactions

Reduction: 2 ClO3^- + 10e- → Cl2

ClO3^- + e- → ClO2

2 Cl- + 2ClO3^- +8e- →2Cl2

Oxidation: 2Cl- → Cl2 + 2e-

Cl- → ClO2 + 5e-

Cl- +ClO3^- → 2ClO2 + 4e-

Step 2: Balance the reactions

2Cl- + 2ClO3^- + 8e- + 12H+ → 2Cl2 + 6H2O

2* (Cl- + ClO3^- + H2O → 2ClO2 + 4e- + 2 H+)

Step 3: The netto reaction

2ClO3^- + 2Cl- + 4H+ → Cl2 + 2H2O +2ClO2

<em>(e) Mn^2 + BiO3^− </em>→<em> Bi^3 + MnO^4− (in acidic solution)</em>

Step 1: The half reactions

Reduction: BiO3^- + 2e- → Bi^3+

Oxidation : Mn^2+ → MnO4^- +5e-

Step 2: Balanced the reactions

5* ( BiO3^- + 2e- + 6H+ → Bi^3+ + 3H2O)

2* ( Mn^2+ + 4H2O →MnO4^- + 5e- + 8H+)

Step 3: The netto reaction

5 BiO3^- + 14 H+ + 2Mn^2+ → 5Bi^3+ + 7H2O + 2MnO4^-

6 0

4 years ago