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Alenkinab [10]
1 year ago
13

How long will it take 10.0 mL of Ne gas to effuse through a porous barrier if it has been observed that 125 minutes are required

for 10.0 mL of Ar gas to effuse through the same barrier
Chemistry
1 answer:
cupoosta [38]1 year ago
7 0

Answer:

88.8 minutes

Explanation:

Graham's law of diffusion relates rate of difusion by the following formula

Rate1 / rate 2 = √( Mass of argon / Mass of Neon)

Where rate = volume divided by time

Rate 1 = 10 ml / t1

Rate 2 = 10 ml / t2

Rate 1/ rate 2 = 10 ml / t1 ÷ 10 ml/ t2 = t2/ t1

t2/t1 = √(Mass of argon / mass of Neon) = √( 39.984/20.179)

125 / t1 = 1.4026

t1 = 125 / 1.4026 = 88.8 minutes

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How do acids and bases affect molecules such Proteins? ​
Taya2010 [7]

Answer:

Strong acids and bases both denature proteins by severing disulphide bonds and at higher temperatures, can break proteins into peptides, or even individual amino acids.

5 0
2 years ago
5. Hydrochloric Acid + Sodium Hydroxide à Sodium<br> Chloride +
erma4kov [3.2K]

HCl Acid + Sodium Hydroxide ----> Sodium  Chloride + water.

<u>Explanation</u>:

  • The reaction between an acid and a base is known as a neutralization reaction. The reaction of an acid with a base to give salt, water and heat is called neutralization.
  • When hydrochloric acid reacts with sodium hydroxide, sodium chloride and water are produced.  

                             HCl + NaOH → NaCl + H2O + Heat

  • The heat evolved in the neutralization reaction raises the temperature of the reaction mixture.    
  • An electro-electrodialysis process (EED) is utilized to create HCl and NaOH from exchange NaCl. NaOH and HCl arrangements with purity higher than 99.9% are acquired. The experimental estimations of the transitions for HCl and NaOH are contrasted and values determined from the incorporation of the Nernst–Planck electro dispersion conditions.
5 0
1 year ago
A 5.00 L sample of helium expands to 12.0 L at which point the
mina [271]

Answer:

1.73 atm

Explanation:

Given data:

Initial volume of helium = 5.00 L

Final volume of helium = 12.0 L

Final pressure = 0.720 atm

Initial pressure = ?

Solution:

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

P₁ × 5.00 L = 0.720 atm × 12.0 L

P₁ = 8.64 atm. L/5 L

P₁ = 1.73 atm

7 0
1 year ago
Which of the following species contains manganese with the highest oxidation number?
ioda

In NaMnO₄, Mn has the highest oxidation number.

The question is incomplete, the complete question is;

Which of the following species contains manganese with the highest oxidation number?

A) Mn

B) MnF₂

C) Mn₃(PO₄)₂

D) MnCl₄

E) NaMnO₄

In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.

1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.

2) For MnF₂;

Mn has an oxidation number of +2

3) For Mn₃(PO₄)₂

Mn has an oxidation number of +2

4) For MnCl₄

Mn has an oxidation number of +4

5) For NaMnO₄

Mn has an oxidation number of +7

Hence in NaMnO₄, Mn has the highest oxidation number.

Learn more: brainly.com/question/10079361

7 0
1 year ago
The combustion of propane is represented by the following chemical equation. C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l) The standard enthalp
wariber [46]

Answer:

ΔH°c = -2219.9 kJ

Explanation:

Let's consider the combustion of propane.

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)

We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.

ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]

ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]

ΔH°c = -2219.9 kJ

7 0
2 years ago
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