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2 years ago

Which is a property of most nonmetallic solids?

Chemistry

2 answers:

dem82 [27]2 years ago

8 0

Answer : The correct option is, (3) brittleness

Explanation :

The properties of nonmetallic solids are :

Nonmetallic solids are brittle in nature.
They are non-malleable and ductile.
Nonmetallic has low thermal conductivity. That means they are poor conductor of electricity.
Nonmetallic has low electrical conductivity. That means they are poor conductor of heat.
They are non-lustrous.
The melting point and boiling point of non metals are lower than the metals.

Hence, the brittleness is a property of most nonmetallic solids

aksik [14]2 years ago

7 0

Most nonmetallic solids have the property of brittleness.

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Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O

Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

ΔH = -793,6 kJ

I hope it helps!

5 0

2 years ago

A gas sample enclosed in a rigid metal container at room temperature (20.0∘C∘C) has an absolute pressure p1p1p_1. The container

choli [55]

Answer: p2 = 1.06p1

Explanation: pressure increases with temperature increase.

According to Gass law

P1/T1 = P2/T2

T1 = 20°c = 20 +273 = 293k

T2 = 40°c = 40 +373 = 313k

Therefore

P2 = P1T2/T1 = 313P2/293

P2 = 1.06P1

3 0

2 years ago

Complete the equation for the dissociation of cdcl2(aq). omit water from the equation because it is understood to be present.

mixas84 [53]

Cadmium chloride is a highly soluble compound. The equation for its dissolution is:

CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)

This dissociation in water allows for the cadmium and chlorine ions to take part in reactions. This is the reason that solutions of chemicals are prepared when a reaction needs to take place.

3 0

2 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

2 years ago

I am going to build a boat. My friend says I cannot put a refrigerator and a television in my boat because that would make it to

Alina [70]

Electric Shock is my best guess.

4 0

2 years ago