CHEMISTRY HIGH SCHOOL

6) At elevated temperatures, methylisonitrile (CH3NC) isomerizes to acetonitrile (CH3CN):CH3NC (g)  CH3CN (g)At the start of an experiment, there are 0.200 mol of reactant and 0 mol of product in the reaction vessel.After 25 min, 0.108 mol of reactant (CH3NC) remain. There are mol of product (CH3CN) in thereaction vessel.A) 0.200 B) 0.540 C) 0.022 D) 0.092 E) 0.308

Answers

Answer 1
Answer:

Answer: The moles of product present in the vessel is 0.092 moles

Explanation:

For the given chemical equation:

CH_3CN(g)\rightleftharpoons CH_3CN(g)

We are given:

Initial moles of reactant = 0.200 moles

Unreacted moles of reactant = 0.108 moles

Reacted moles of reactant = 0.200 - 0.108 = 0.092 moles

By Stoichiometry of the reaction:

1 mole of reactant produces 1 mole of product

So, 0.092 moles of reactant will produce = \frac{1}[1}* 0.092=0.092mol of product

Hence, the moles of product present in the vessel is 0.092 moles

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What Kelvin temperature is equal to -33DC? A) 306K B) -33K C) 240K D) 33K

Answers

Final Answer:

\boxed{\boxed{\huge\textsf{$\rm -33^(\,\circ)C=240.15\,K$}}}

Kelvin Absolute Temperature:

The kelvin, symbol K, is a unit of measurement for temperature. The Kelvin scale is an absolute scale, which is defined such that 0 K is absolute zero, the lowest temperature possible on Earth.

Historically developed from the Celsius scale, the conversion is thus:

\boxed{\Large\textsf{$\rm 0^(\,\circ)C=273.15\,K$}}

And absolute zero, 0 K = -273.15° C, the lowest temperature possible.

Knowing this, we can solve the question as:

\Large\textsf{$\rm -33^(\,\circ)C=(-33+273.15)\,K$}\n\n\boxed{\boxed{\Large\textsf{$\therefore\rm -33^(\,\circ)C=240.15\,K$}}}

\hrulefill

To learn more about the Kelvin Absolute Temperature Scale:

brainly.com/question/24652526
1 ℃ = 273.15 K.

The Kelvin temperature equal to -33℃ is 240 K.

What are the 3 parts of a nucleotide

Answers

The 3 parts are

1) A five carbon ribose sugar

2) A Phosphate molecule

3)  The four nitrogenous bases


I hope that's help ! 
Like DNA, RNA polymers are make up of chains of nucleotides. These have three parts a five carbon ribose sugar. A phosphate molecule . And one of four nitrogenous bases adenine, guanine, cytosine, or uracil

What is electron configuration of sulfer?

Answers

Sulfur has 16 protons and 16 electrons.
The electronic configuration of Sulfur is:
2, 8, 6        =          16 

If 20.0 ml of glacial acetic acid (pure hc2h3o2 is diluted to 1.70 l with water, what is the ph of the resulting solution? the density of glacial acetic acid is 1.05 g/ml.

Answers

The pH of the resulting solution is 2.72

Data;

  • volume of acid = 20mL
  • density of acid = 1.05 g/ml
  • mass = 1.05 * 20  = 21g

pH of a solution

The number of moles of acetic acid is

n = 21/60 = 0.333 moles

The concentration of acetic acid in 1.7L is

c = (0.333)/(1.7) = 0.196M

The equation for dissociation is

CH_3COOH + H_2O \to CH_3COO^- +H_3O^+

The Ka of this reaction is

K_a = ([CH_3COO^-][H_3O^+)/([CH_3COOH)

The Ka of acetic acid = 1.8*10^-5

1.8*10^-^5 = (x*x)/(0.196) \nx = 0.187*10^-^2M\n

The concentration of hydrogen ion is

[H^+] = 1.87*10^-^3M

The pH of this solution is

pH = -log[H^+]\npH = -log(1.87*10^-^3)\npH = 2.72

The pH of the resulting solution is 2.72

Learn more on pH of a solution here;

brainly.com/question/25639733
Given:
20.0 ml of glacial acetic acid
1.70 L with water
density of the glacial acetic acid is 1.05 g/ml ; Ka = 1.8 x 10⁻⁵

1) find out how many grams of HC2H3O2 is in this glacial matter. It says here it has a density of 1.05g/mL. 
So we find the grams from that => 1.05 g/mL * (20.0 mL) = 21 grams of HC2H3O2. 
Convert this to moles => 21 grams (1 mol / 60 g) = 0.35 mol HC2H3O2 

We then get the Ka formula from that equation so it is 
Ka = [H3O+] [C2H3O2] / [HC2H3O2] (Remember H2O has an activity of 1 so we exclude it) 

We know the concentration of HC2H3O2 because we have moles of HC2H3O2 
Finding concentration of HC2H3O2 => 0.35 mol HC2H3O2/1.70 L = 0.20588 M - initial concentration

1.8 x 10⁻⁵ = (x) (x) / (0.20588 M - x)  
1.8 X 10⁻⁵ (0.20588 M) = x²
3.70584⁻ x 10⁻⁶ = x²
x = 1.9251 x 10⁻³

pH = -log (H₃O⁺) = - log (1.9251 x 10⁻³) = -(-2.7156) = 2.7156

pH of the resulting solution is 2.7156 

A 25°C sample of gas has a volume of 5.0 L. What will it’s volume be if it is heated to 50.0°C?

Answers

The volume of the gas will be 0.8385 when heated to 50.0°C

Which chemical equation best describes the reaction of potassium metal with fluorine gas to form solid potassium fluoride? A. K(s) + F2(g) → KF2(s)

B. K(s) + F(g) → KF(s) 

C. K2(g) + F2(g) → 2 KF(s)

D. 2 K(s) + F2(g) → 2 KF(s) 

I guessed B

Answers

A. K(s) + F2(g) → KF2(s) 

B. K(s) + F(g) → KF(s) 

C. K2(g) + F2(g) → 2 KF(s) 

D. 2 K(s) + F2(g) → 2 KF(s) 

 it exists as a pale yellow diatomic gas at standard conditions!!
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