CHEMISTRY COLLEGE

Write the balanced chemical equation for each of these reactions. Include phases.1) When aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms.

2) However, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble Pb(OH)42-(aq) complex ion

Answers

Answer 1

Answer:

Answer: The chemical equations are given below.

Explanation:

For 1:

The chemical equation for the reaction of lead nitrate and sodium hydroxide follows:

$Pb(NO_3)_2(aq.)+2NaOH(aq.)\rightarrow Pb(OH)_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

1 mole of aqueous solution of lead nitrate reacts with 2 moles of aqueous solution of sodium hydroxide to produce 1 mole of solid lead hydroxide and 2 moles of aqueous solution of sodium nitrate.

For 2:

The chemical equation for the reaction of lead hydroxide and hydroxide ions follows:

$Pb(OH)_2(s)+2OH^-(aq.)\rightarrow [Pb(OH)_4]^(2-)(aq.)$

By Stoichiometry of the reaction:

1 mole of lead hydroxide reacts with 2 moles of aqueous solution of hydroxide ions to produce 1 mole of aqueous solution of tetra hydroxy lead (II) complex

Hence, the chemical equations are given above.

Answer 2

Answer:

The balanced chemical equation for each of the reactions is as follows:

Pb(NO3)2(aq) + 2NaOH(aq) → 2 NaNO3(aq) + Pb(OH)2(s)
Pb(OH)2 + 2OH- → Pb(OH)42-

How to balance chemical equation?

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation are the same.

According to this question, when aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms. The balanced equation are as follows:

Pb(NO3)2(aq) + 2NaOH(aq) → 2 NaNO3(aq) + Pb(OH)2(s)

Also, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble Pb(OH)42-(aq) complex ion. The balanced equation is as follows:

Pb(OH)2 + 2OH- → Pb(OH)42-

Learn more about balanced equation at: brainly.com/question/1301642

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Be sure to answer all parts. (a) How many atoms are directly bonded to the central atom in a trigonal planar molecule?

i. two
ii. three
iii. six
iv. eight

(b) How many atoms are directly bonded to the central atom in a trigonal bipyramidal molecule?

i. three
ii. four
iii. five
iv. six

(c) How many atoms are directly bonded to the central atom in an octahedral molecule?

i. three
ii. four
iii. six
iv. eight

Answers

Answer:

a) ii

b)iii

c)iii

Explanation:

three atoms directly bonded then only it is possible to achieve trigonal planar

trigonal bipyramidal means five atoms should attach to central atom

for octahedral six atoms must directly connected to central atom

You have been injured in the laboratory (cut,burn,etc) first you should

Answers

Let the teacher know

Answer:tell the instructor of the lab

Explanation:

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first ignoring ionic strength and activities. a. silver iodate
b. barium sulfate
c. Repeat the above calculations using ionic strength and activities.

Answers

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_(sp) =\text{[Ag$^(+)$][IO$_(3)$$^(-)$]} = s* s = s^(2) = 3.0* 10^(-8)\ns = \sqrt{3.0* 10^(-8)} \text{ mol/L} = 1.7 * 10^(-4) \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_(sp) =\text{[Ba$^(2+)$][SO$_(4)$$^(2-)$]} = (0.02 + s) * s \approx 0.02s = 1.1* 10^(-10)\ns = (1.1* 10^(-10))/(0.02) \text{ mol/L} = 5.5 * 10^(-9) \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = (1)/(2) \sum_(i) {c_(i)z_(i)^(2)}\n\n\mu = (1)/(2) (\text{[Ba$^(2+)$]}\cdot (2+)^(2) + \text{[NO$_(3)$$^(-)$]}*(-1)^(2)) = (1)/(2) (\text{0.02}* 4 + \text{0.04}*1)= (1)/(2) (0.08 + 0.04)\n\n= (1)/(2) *0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(1)^(2)√(0.06) = -0.51* 0.24 = -0.12\n\gamma = 10^(-0.12) = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_(sp) =\text{[Ag$^(+)$]$\gamma_(Ag^(+))$[IO$_(3)$$^(-)$]$\gamma_{IO_(3)^(-)}$} = s*0.75* s * 0.75 =0.56s^(2)= 3.0 * 10^(-8)\ns^(2) = (3.0 * 10^(-8))/(0.56) = 5.3 * 10^(-8)\n\ns =2.3 * 10^(-4)\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(2)^(2)√(0.06) = -0.51*16* 0.24 = -0.50\n\gamma = 10^(-0.50) = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_(sp) =\text{[Ba$^(2+)$]$\gamma_( Ba^(2+))$[SO$_(4)$$^(2-)$]$\gamma_{ SO_(4)^(2-)}$} = (0.02 + s) * 0.32* s* 0.32 \approx 0.02*0.10s\n2.0* 10^(-3)s = 1.1 * 10^(-10)\ns = (1.1* 10^(-10))/(2.0 * 10^(-3)) \text{ mol/L} = 5.5 * 10^(-8) \text{ mol/L}$

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 and

liters of the 25% solution must be mixed to obtain a 28% solution of H2SO4.

(Round to the nearest tenth, if necessary.)

Answers

Answer:

16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

Explanation:

Let the required volume of solution 1 be represented by x.

The required volume of solution 2 would then be 78-x.

The number of moles of solution 1 that would be required = 0.4x

The number of moles of solution 2 that would be required = 0.25(78-x)

The number of moles of the final mixture = 78 x 0.28 = 21.84

moles of solution 1 + moles of solution 2 = moles of final mixture

0.4x + 0.25(78 - x) = 21.84

0.4x + 19.5 - 0.25x = 21.84

0.4x - 0.25x = 21.84 - 19.5

0.15x = 2.34

x = 15.6 liters

To the nearest tenth = 16 liters

Liters of 40% solution needed = 16 liters

Liters of 25% solution needed = 78 - 16 = 62 liters.

Hence, 16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

A chemist prepares a solution of magnesium fluoride MgF2 by measuring out 0.00598μmol of magnesium fluoride into a 50.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's magnesium fluoride solution. Round your answer to 2 significant digits.

Answers

Answer:

0,12 μmol/L of MgF₂

Explanation:

Preparation of solutions is a common work in chemist's life.

In this porblem says that you measure 0,00598 μmol of MgF₂ in 50,0 mL of water and you must calculate concentration in μmol/L

You have 0,00598 μmol but not Liters.

To obtain liters you sholud convert mL to L, knowing 1000mL are 1 L, thus:

50,0 mL (1L/1000mL) = 0,05 L of water.

Thus, concentration in μmol/L is:

0,00598 μmol / 0,05 L = 0,12 μmol/L -The problem request answer with two significant digits-

I hope it helps!

Lithium (Li) has a charge of +1, and oxygen has a charge of -2. Which is the chemical formula?

Answers

Considering the formation of a chemical formula, the chemical formula is Li₂O.

Ionic compounds

Cations (positivelycharged ions) and anions (negatively charged ions) combine to form ionic compounds, which must be electrically neutral. Therefore, the cations and anions must combine in such a way that the net charge contributed by the total number of cations exactly cancels the net charge contributed by the total number of anions.

Formation of a chemical formula

To form the chemical formula:

It is first necessary to place the chemical symbol of the element that is a cation, and then the negative part (anion).
Since the net charge of the compound formed must be zero (that is, add zero), when cations and anions with different charges are joined, it is necessary to cross them and place them as subscripts of the other element. In this way, balanced loads are obtained, that is, the algebraic sum is equal to zero.

Chemical formula in this case

Lithium (Li) has a charge of +1, and oxygen has a charge of -2. Taking into account the above, the chemical formula is Li₂O.

Answer:

Lithium formula=Li+

Oxygen formula=O2(2-)

Explanation:

Quick note: These kinds of formula are really easy to google. Next time, google the chemical name, include the charge and include "formula" and you should get the answer.

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