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ale4655 [162]
3 years ago
12

Ionic character of lih

Chemistry
1 answer:
adoni [48]3 years ago
8 0
I believe LiH becomes HF 
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For an alloy that consists of 93.1 g copper, 111.7 g zinc, and 4.0 g lead, what are the concentrations of (a) Cu, (b) Zn, and (c
Andrew [12]

Answer:

(a) weight percent of Cu = 44.59%

(b) weight percent of Zn = 53.49%

(c) weight percent of Pb = 1.91%

Explanation:

Given: Alloy contains- Cu=93.1 g, Zn=111.7 g, Pb=4.0 g

Therefore, the total mass of the alloy (m) = mass of Cu (m₁) + mass of Zn (m₂)+ mass of Pb (m₃)

m= 93.1 g + 111.7 g + 4.0 g = 208.8 g

(a) weight percent of Cu = (m₁ ÷ m)× 100% =  (93.1 g ÷ 208.8 g)× 100% =44.59%

(b) weight percent of Zn = (m₂ ÷ m)× 100% =  (111.7 g ÷ 208.8 g)× 100% =53.49%

(c) weight percent of Pb = (m₃ ÷ m)× 100% =  (4.0 g ÷ 208.8 g)× 100% =1.91%

5 0
3 years ago
Which statments regarding the henderson-hasselbalch equation are true?
ziro4ka [17]

Complete question is;

Which statements regarding the Henderson-Hasselbalch equation are true?

1. If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined.

2. At pH = pKa for an acid, [conjugate base] = [acid] in solution.

3. At pH > pKa for an acid, the acid will be mostly ionized.

4. At pH < pKa for an acid, the acid will be mostly ionized.

A. All of the listed statements are true. B. 1, 2, and 3 are true.

C. 2, 3, and 4 are true.

D. 1, 2, and 4 are true.

Answer:

B. 1, 2, and 3 are true.

Explanation:

The formula for the Henderson-Hasselbalch equation is:

pH = pka + log₁₀([A^(-)]/[HA])

Where;

PH is acidity of solution

ka is acid dissociation constant

A^(-) is concentration of conjugate base

HA is concentration of Acid

- For statement 1; If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined;

pH = pka + log₁₀([A^(-)]/[HA])

pH - pka = log₁₀([A^(-)]/[HA])

10^(pH - pka) = ([A^(-)]/[HA])

Since we can find the ratio as seen, then the statement is true

- For statement 2: At pH = pKa for an acid, [conjugate base] = [acid] in solution;

We will substitute pH for pKa;

pH = pH + log₁₀([A^(-)]/[HA])

This give;

0 = log₁₀([A^(-)]/[HA])

10^(0) = [A^(-)]/[HA]

1 = [A^(-)]/[HA]

Thus; [A^(-)] = [HA]

Thus, the statement is true

- For statement 3: At pH > pKa for an acid, the acid will be mostly ionized;

This means that;

pH - pKa is greater than 0 and thus;

10^(pH - pKa) is greater than 1.

Thus;

[A^(-)]/[HA] > 1

[A^(-)] > [HA]

So more acid is ionized than base.

So the statement is true.

- For statement 4: At pH < pKa for an acid, the acid will be mostly ionized;

This means that;

pH - pKa is less than 0 and thus;

10^(pH - pKa) is less than 1.

Thus;

[A^(-)]/[HA] < 1

[A^(-)] < [HA]

So we have more base ionized than acid. So statement is false

7 0
3 years ago
An atom has mass number 23 and atomic number 11.
aleksklad [387]

A): The closest orbital to the nucleus, called the 1s orbital, can hold up to two electrons. This orbital is equivalent to the innermost electron shell of the Bohr model of the atom. It is called the 1s orbital because it is spherical around the nucleus. The 1s orbital is always filled before any other orbital.

6 0
2 years ago
Read 2 more answers
The addition of nutrients to<br> water by human activity is called<br> artificial
nikitadnepr [17]

Answer:

The addition of nutrients to water by human activity is called artificial Eutrophication. Nonpoint-source-pollution comes from many small sources.

Explanation:

hope this helps :)

5 0
3 years ago
How many grams are in 1.11 moles of MN3(SO4)7
evablogger [386]
Lemme look at the periodic table give me 1 minute

7 0
3 years ago
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