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luda_lava [24]
3 years ago
10

How to prepare 100 mL of lead nitrate Pb(NO3)2 ?

Chemistry
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer:

It is obtained by dissolving lead(II) oxide in nitric acid

Explanation:

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Ephedrine, a central nervous system stimulant, is used in nasalsprays as a decongestant. this compound is a weak organic base: {
sashaice [31]

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base:

C10H15ON (aq) + H2O (l) -> C10H15ONH+ (aq) + OH- (aq)

A 0.035 M solution of ephedrine has a pH of 11.33.

a) What are the equilibrium concentrations of C10H15ON, C10H15ONH<span>+, and OH-</span>?

b) Calculate <span>Kb</span> for ephedrine.

c(C₁₀H₁₅NO) = 0,035 M.<span>
pH = 11,33.
pOH = 14 - 11,33 = 2,67.
[OH</span>⁻] = 10∧(-2,67) = 0,00213 M.<span>
[OH</span>⁻] = [C₁₀H₁₅NOH⁺] = 0,00213 M.<span>
[</span>C₁₀H₁₅NO] = 0,035 M - 0,00213 M = 0,03287 M.<span>
Kb = [OH</span>⁻] · [C₁₀H₁₅NOH⁺] / [C₁₀H₁₅NO].<span>
Kb = (</span>0,00213 M)² / 0,03287 M = 1,38·10⁻⁴.

3 0
3 years ago
Isotopes are atoms of an element
stellarik [79]
D. With the same number of protons and different number of neutrons.
6 0
3 years ago
There are ________ unpaired electrons in the Lewis symbol for a nitride ion.
Anna71 [15]

Answer:

There are no unpaired electrons.

Explanation:

There are  no unpaired electrons in the Lewis symbol for a nitride ion({ N }^{ 3- }).The nitride ion has a charge of -3. The negative charge on the Nitride ion indicates a gain in electrons . Nitrogen has 5 valence electrons that is the number of electrons that are in its outer shell .The total number of electrons that the nitride ion has is equal to 5+3 = 8 electrons . Electrons usually appear in pairs and obey the octet rule therefore the nitride ion has four electron pairs no unpaired electrons.

5 0
3 years ago
Using the Brønsted-Lowry concept of acids and bases, identify the Brønsted-Lowry acid and base in each of the following reaction
Luden [163]

Answer:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

<u>The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻</u>

<u />

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

<u>There are no Brønsted-Lowry acids and bases in this reaction.</u>

Explanation:

According to the Brønsted-Lowry concept, when an acid (HA) and a base (B) undergoes a chemical reaction, the acid (HA) loses a proton and forms its conjugate base (A⁻), whereas the base gains (B) the proton to form its conjugate acid (HB⁺).

<em>The chemical equation for this reaction is:</em>

HA  +  B  ⇌  A⁻  +  HB⁺

Given reactions:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

<u>The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻</u>

Reason: In this reaction, the acid H₂O loses a proton and forms its conjugate base, OH⁻. Whereas, the base HSO³⁻ gains a proton to form its conjugate acid, H₂SO₃.

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

<u>There are no Brønsted-Lowry acids and bases in this reaction.</u>

Reason: In this reaction, there is no exchange of proton between the acid and the base.

4 0
3 years ago
What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
Vikentia [17]

Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

4 0
3 years ago
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