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Lynna [10]
2 years ago
11

Differentiate between positive and negative deviation ~

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ty! :)​
Chemistry
1 answer:
Montano1993 [528]2 years ago
7 0

Answer:

positive deviation from raoults law means when the vapour pressure is higher than the solution is expected to exhibit e.g of one is the mixture of acetone and ethanol

negative deviation means that it is opposed to the total vapour pressure of the solution

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It took 55 days for a radioactivity of 1.75 x 1012 Bq to remain 0.135 Ci. What is the half-life of this radioactivity?
Mnenie [13.5K]

From the calculations, the half life of the material is 6.5 days.

<h3>What is radioactivity?</h3>

The term radioactivity has to do with the spontaneous disintegration of a specie.

Uisng the formula;

N=Noe^-kt

N= amount at time t = 0.135 Ci or 4.995 ×10^9 Bq

No = amount initially present =  1.75 x 10^12 Bq

k = rate constant = ?

t = time taken = 55 days

Hence;

4.995 ×10^9  = 1.75 x 10^12e^-55k

4.995 ×10^9/1.75 x 10^12 = e^-55k

2.85 * 10^-3 = e^-55k

ln2.85 * 10^-3 = -55k

k = ln2.85 * 10^-3/-55

k = 0.1066 day-1

Half life = 0.693/ 0.1066 day-1

= 6.5 days

Learn more about radioactivity:brainly.com/question/1770619

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4 0
1 year ago
Which atom has the least attraction for the electrons in a bond between that atom and an atom of hydrogen?
Helga [31]

Answer:

carbon

Explanation:

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8 0
3 years ago
Why is it difficult to accurately represent ammonia and methane on paper?
andrew-mc [135]

Answer:

The correct answer is because the molecular structure.

Explanation:

The difficulty of ammonia and methane to be represented on paper is due to the molecular structure. These compounds have a three-dimensional projection with defined angles. Ammonia presents angles of 109.5º between the atom of Nitrogen and those of Oxygen. The ammonia presents 107.8º between the oxygen atoms.

In the methane molecule, there is 109.5º between the hydrogen molecules and the carbon atom. This results in the need for a 3D representation of the molecule.

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6 0
3 years ago
How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant
schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

  • 1 g = 1,000 mg
  • The molar mass of Cu is 63.55 g/mol
  • When 1 mole of Cu is deposited, 2 moles of electrons circulate.
  • The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
  • 1 A = 1 C/s

The time  that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s

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