Chlorine (Cl2Cl2) is a strong germicide used to disinfect drinking water and to kill microbes in swimming pools. If the product is Cl−Cl−, was the elemental chlorine oxidized or reduced?

Answers

Answer 1

Answer:

The elemental chlorine was reduced

Explanation: inl the elemental form, the oxidation state of chlorine is =0 on been reduced to Cl-Cl-, the oxidation state reduces to -1

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If the partial pressure of N2 in a scuba divers blood at the surface is 0.79 atm, what will the pressure be if he/she descends to a depth of 30 meters (4 atm) and stays there long enough to reach equilibrium (b)

Answers

Answer:

Explanation:

for every 3m that the internal pressure is lowered, it increases in an atmosphere approximately, so when the blood pressure of nitrogen decreases 30m, it will increase by approximately 10 atm, being enough there for the body to enter into equilibrium

Select the complexes that exhibit geometric isomerism. check all that apply. check all that apply. [co(h2o)2(ox)2]− [co(h2o)2(nh3)2(ox)]+ [co(en)3]3+ [ni(co)2cl2] [ni(nh3)2(en)2]2+

Answers

Geometric isomerism in another name is called cis-trans isomerism or it is called E-Z isomerism.
[Co(H2O)2(ox)2]- it is an octahedral complex which has two of bidentate ligand and another two of ligand. It exists as cis-trans isomers.
[Co(en)3]³+ it is an octahedral complex which has three bidentate ligands and it cannot exist as cis-trans isomers.
[Ni(CO)2CL2] is a square planar complex which has two of one ligand another two of ligand. It exists as cis-trans isomers.
[Co(H2O)2 (NH3)2(ox)]+ it is an octahedral complex with one of bidentate ligand two of each another ligand. It exists as cis-trans isomers.
[Ni(NH3)2 (en)2]²+ is an octahedral complex with one of a bidentate ligand and two each of another ligand. It exists as cis-trans isomers.
Following compounds exhibit geometric isomerism.
[Co(H2O)2 (ox)2]-
[Co(H2O2) (NH3)2 (ox)]+
[Ni(Co)2 CL2]
[Ni(NH3)2 (en)2]²+

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH solution is added, the pH in the titration flask will be a. 2.17
b. 3.35
c. 2.41
d. 1.48
e. 7.00

Answers

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

Final answer:

The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.

Explanation:

The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.

First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.

Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].

To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.

Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.

Learn more about titration here:

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Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-chlorobutane. Based on this information, the major organic product(s) of this reaction are expected to be _____.

Answers

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

Ri1. Among these which is the example of compound?
a) sulphur
b) iron
c) ammonia
der

Answers

Answer:

Ammonia is a compound.

Explanation:

A compound is composed of two or more separate elements. Ammonia is NH3, which happens to be two different elements.

Which words or phrases identify the types of temperate climates? Check all that apply. humid continental

highland
it is actually science on the subject but it doesn't have that option.
marine west coast

Mediterranean

subarctic

tropical wet-dry

Answers

Marine west coast and Mediterranean are the types of temperate climates, due to the dispersion of precipitation throughout the year, temperate marine climates are typically distinguished by a notable lack of dry season, hence options D and E are correct.