CHEMISTRY COLLEGE

Of the following, which element has the highest first ionization energy?Li
Cs
At
F​

Answers

Answer 1
Answer: There are free tutor on my website just got to freetutor.com
Answer 2
Answer: I think it’s Li but I don’t totally know

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A chemist has two solutions of H2SO4. One has a 40% concentration and the other has a 25% concentration.How many liters of each solution must be mixed to obtain 78 liters of a 28% solution?liters of the 40% solution andliters of the 25% solution must be mixed to obtain a 28% solution of H2SO4.(Round to the nearest tenth, if necessary.)

Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of Ni and H+ in the balanced reaction? Ni2+(aq) + NH4+(aq) → Ni(s) + NO3-(aq)

Answers

In this case, the problem is asking for the balance of a redox reaction in acidic media, in which nickel is reduced to a metallic way and nitrogen oxidized to an ionic way.

Thus, according to the given information, it turns out possible for us to balance this equation in acidic solution by firstly setting up the half reactions:

Ni^(2+)+2e^-\rightarrow Ni^0\n\nN^(3-)H_4^++3H_2O\rightarrow N^(5+)O_3^-+8e^-+10H^+

Next, we cross multiply each half-reaction by the other's carried electrons:

8Ni^(2+)+16e^-\rightarrow 8Ni^0\n\n2N^(3-)H_4^++6H_2O\rightarrow 2N^(5+)O_3^-+16e^-+20H^+

Finally, we add them together to obtain:

8Ni^(2+)+2N^(3-)H_4^++6H_2O\rightarrow 8Ni^0+2N^(5+)O_3^-+20H^+

Which can be all simplified by a factor of 2 to obtain:

4Ni^(2+)+N^(3-)H_4^++3H_2O\rightarrow 4Ni^0+N^(5+)O_3^-+10H^+\n\n4Ni^(2+)(aq)+NH_4^+(aq)+3H_2O(l)\rightarrow 4Ni(s)+NO_3^-(aq)+10H^+(aq)

Hence, the coefficients in front of Ni and H⁺ are 4 and 10 respectively.

Learn more:

.Which statement describes the difference between incomplete dominance and codominance?In codominance, only one allele is expressed in the offspring; In incomplete dominance, both alleles are expressed in the offspring.
In codominance, both allels are expressed in the offspring; In incomplete dominance, only one allele is expressed in the offspring.
In codominance, both alleles are expressed in the offspring; in incomplete dominance, the offspring demonstrate an intermediate form of the alleles from the parents.
In codominance, the offspring demonstrate an intermediate form of the alleles from the parents; in incomplete dominance, both alleles are expressed in the offspring.

Answers

Answer:

C Because im in 3rd grade and just did the test

Explanation:

im smart im in third grade but your doing this in college

Answer:

answer is c: In codominance, both alleles are expressed in the offspring; in incomplete dominance, the offspring demonstrate an intermediate form of the alleles from the parents.

Explanation:

given that the only known ionic charges of lead are pb(ii) and pb(iv), how can you explain the existence of the pb2o3 salt

Answers

Answer:

See Explanation

Explanation:

Pb2O3 is better formulated as PbO.PbO2. It is actually a mixture of the two oxides of lead, lead II oxide and lead IV oxide.

This implies that this compound Pb2O3  (sometimes called lead sesquioxide) is a mixture of the oxides of lead in its two known oxidation states +II and +IV.

Hence Pb2O3  contains PbO and PbO2 units.

A chemist prepares hydrogen fluoride by means of the following reaction:CaF2 + H2SO4 --> CaSO4 + 2HF

The chemist uses 11 g of CaF2 and an excess of H2SO4, and the reaction produces 2.2 g of HF.
(a) Calculate the theoretical yield of HF.

(b) Calculate the percent yield of HF.

Answers

Answer:

39.3%

Explanation:

CaF2 + H2SO4 --> CaSO4 + 2HF

We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:

For CaF2;

Number of moles reacted= mass/molar mass

Molar mass of CaF2= 78.07 g/mol

Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride

Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride

0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride

Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass

Molar mass of hydrogen fluoride= 20.01 g/mol

Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)

Actual yield of HF was given in the question as 2.2g

% yield of HF= actual yield/ theoretical yield ×100

%yield of HF= 2.2/5.6 ×100

% yield of HF= 39.3%

An aqueous solution is 4.44 M nitric acid and the density of the solution is 1.42 g/mL. Calculate the mole fraction of this solution.

Answers

The mole fraction of HNO3 is  0.225

Explanation:

1.Given data

Density = 1.429 /ml

Mass% = 63.01 g HNO3 / 100g of solution

The mass of 63.01 g is in 100 / 1.142 /ml of solution

Or 63.01 g in 55.7 mL

Molarity = 15.39 moles / L

Mass of water in 100g = 100 - 63.01=36.99 g

So 63.01 grams in 36.99 grams of water

So mass of HNO3 in 1000grams of water = 63.01* x 1000 / 36.99 = 1703

Moles of HNO3 in 1000g = 1703 / 63.01 = 27.03 moles

Molality = 27.03 molal (mole / Kg)

Mole fraction = Mole of HN03 / Moles of water + mole of HNO3

Mole of water = 62/ 18 = 3.44

Moles of HNO3 = 63.01 / 63.01 = 1.000

Mole fraction = 1.000 / 3.44 + 1.000 = 0.225

The mole fraction of HNO3 is  0.225

1. The common name for the compound CH3-CH2-O-CH3 is ​

Answers

Answer:

propane

Explanation:

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Answer:

Methoxyethane also known as ethyl methyl ether
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