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Degger [83]
3 years ago
14

A student comes to the conclusion that solids are denser than liquids. Is this true? Explain.

Chemistry
1 answer:
SashulF [63]3 years ago
6 0
True, because a liquid can be taken and added, but a solid stay the same never losses and never gains
You might be interested in
16. A sample of nitrogen gas,
Lana71 [14]

find mol of N2 present using gas law equation

PV = nRT

P = pressure = 688/760 = 0.905 atm.

V = 100mL = 0.1L

n = ???

R = 0.082057

T = 565+273 = 838

Substitute:

0.905*0.1 = n*0.082057*838

n = 0.0905 / 68.76

n = 0.00132 mol N2

Molar mass N2 = 28 g/mol

0.00132 mol = 0.00132*28 = 0.037g N2 gas

4 0
3 years ago
Conduct metric Titration of H_2(SO_4) and Ba(OH)_2 Write an equation (including states of matter) for the reaction between H_2(S
meriva

Answer:

a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)

b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.

c) H⁺, HSO₄⁻, SO₄²⁻

d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.

f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Explanation:

a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.

The <em>balanced equation</em> is:

H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)   [1]

b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?

In the beginning there is H₂SO₄ and the ions that come from its <em>dissociation reactions</em>: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its <em>ionization</em>.

H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)

H₂O(l)  ⇄ H⁺(aq) + OH⁻(aq)

c) What is the conducting species in this initial solution?

The main responsible for conductivity are the <em>ions</em> coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.

d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?

As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?

At the <em>equivalence point</em> equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are <em>weak electrolytes</em>, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.

f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

7 0
3 years ago
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
For each of the bonds listed, tell which atom is more negatively charged a. C--H b. C--N c. C--S d. C--O
blondinia [14]

Explanation:

The atom is considered to be more electronegative , when it has the tendency to get a slight negative i.e. , delta negative charge on it , and the other atom gets a delta positive charge .

In the periodic table ,

As we go left to right in the periodic table , the electronegativity increases ,

and

As we go top to bottom , the electronegativity decreases .

Hence , from the option given in the question ,

a. C--H , the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , Carbon is more electronegative than hydrogen .

b. C--N ,  the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , nitrogen is more electronegative than Carbon .

c. C--S , the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , sulfur is more electronegative than Carbon .

d. C--O , the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , oxygen is more electronegative than Carbon .

5 0
2 years ago
Help please I’ll mark as brainiliest
fredd [130]

Answer:

         B             B

W      BW          BW        

W       BW          BW

Explanation:

Hope this helps

8 0
2 years ago
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