Of Sr or Ba , the element with the higher first ionization energy is

Answers

Answer 1

Answer:

Explanation:

Sr has an ionization of 550 whereas Ba has an ionization of 503

Related Questions

Which of the following would be more reactive than magnesium (Mg)?
A. Calcium (Ca)
B. Potassium (K)
C. Argon (Ar)
D. Beryllium (Be)

Answers

Answer:potassium is more reactive than Mg because both lie in the same group and the element potassium has more electropositivity than magnesium

Explanation:

I hope it will help you

Answer: B. Potassium(K)

Explanation:

A piece of wood near a fire is at 23°C. It gains 1,160 joules of heat from the fire and reaches a temperature of 42°C. The specific heat capacity of
wood is 1.716 joules/gram degree Celsius. What is the mass of the piece of wood?
ОА. 16 g
OB. 29 g
ОC. 36 g
OD. 61 g

Answers

Answer:

35.578g or 36g if you round

Explanation:

Q=mc ∆∅ where ∅ is temperature difference

1160= m x 1.716 x (42-23)

m = 1160/ 1.716 x19

m=35.578g

m = 36g to nearest whole number

Answer: C. 36 g

Explanation: I got this right on Edmentum.

25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311 M NaOH solution. What is the concentration of the H2SO4 solution

Answers

Answer:

Concentration of the H₂SO₄ solution is 0.0737 M

Explanation:

Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:

H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O

From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.

mole ratio of acid to base, nA/nB = 1:2

Concentration of the base, Cb = 0.1311 M

Volume of base, Vb, = 28.11 mL

Concentration of acid, Ca = ?

Volume of acid, Va + 25.0 mL

Using the formula, CaVa/CbVb = nA/nB

making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB

substituting the values into the equation

Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M

Therefore, concentration of the H₂SO₄ solution is 0.0737 M

Given the following Fischer projection: Fischer projection for an entantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right. draw the perspective formula of the molecule. Be sure to correctly indicate stereochemistry in your answer.

Answers

Answer:

Explanation:

Stereoisomers are two or more atoms that have the same bonding order of atoms but there is a difference spatial arrangement of the atoms in space.

A plane of symmetry divides a molecule into two equal halves.

A chiral stereoisomer are not superimposed on a mirror image , Hence they do not posses a plane of symmetry.

As a result to that. these non-superimposable mirror images are said to be Enantiomers.

However, a Fischer Projection emanates from a two - dimensional figure which is used for presenting a three - dimensional organic molecules.

From the given question;

Fischer projection for an enantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right.

we can sketch the way the enantiomer of 2-bromo-2,3-dihydroxypropanal can be seen like the one shown below:

CH₂OH

Br -------------|----------------OH

CHO

The objective of this question is to drawn the perspective formula of the molecule.

So , from the attached file below; we can see the perspective formula of the molecule in a well structured 3-D format.

a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a buffer composed of HC2H3O2 and C2H3O2 - ? Write a chemical equation for that reaction.c) What happens when HBr is added to this buffer? Write a chemical equation for that reaction.

Answers

Answer:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

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