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Stolb23 [73]
3 years ago
15

Which of these can be classified as a pure substance?

Chemistry
2 answers:
Mariana [72]3 years ago
6 0
Salt. It's not mixed with anything.
vredina [299]3 years ago
4 0
Sssaaaallllttttt!!!!!!
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Polyester and Nylon are examples of synthetic materials called polymers.<br> A. True<br> B. False
Rainbow [258]

Answer:

The answer is true

7 0
2 years ago
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What is a mol in chemistry
Snowcat [4.5K]
Molecular weight it stands for molecular weight






6 0
3 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
100 ml is drawn from 0.1 M solution of KCl and added to 900 ml of water. What is the
grandymaker [24]

Answer:

The new concentration will be 0.01 M.

Explanation:

To determine the new concentration we use the following formula.

concentration (1) × volume (1) = concentration (2) × volume (2)

concentration (1) = 0.1 M

volume (1) = 100 mL

concentration (2) = unknown

volume (2) = 100 mL + 900 mL = 1000 mL

concentration (2) = [concentration (1) × volume (1)] / volume (2)

concentration (2) = (0.1 × 100) / 1000 = 0.01 M

3 0
3 years ago
A chemist fills a reaction vessel with mercurous chloride solid, mercury (I) aqueous solution, and chloride aqueous solution at
makvit [3.9K]

Answer:

ΔG° = -533.64 kJ

Explanation:

Let's consider the following reaction.

Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)

The standard Gibbs free energy (ΔG°) can be calculated using the following expression:

ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)

where,

ni are the moles of reactants and products

ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products

ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)

ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)

ΔG° = -533.64 kJ

3 0
3 years ago
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