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lukranit [14]
3 years ago
15

Why must the ph ammonia buffer solution be stored and dispensed in the fume exhaust hood?

Chemistry
1 answer:
Vadim26 [7]3 years ago
7 0

On the other hand ammonia is a very dangerous chemical which has a pungent smell and effect the eyes of the user. Thus it kept always in the fume exhaust hood for storing and dispensing function.  

The  pH of ammonia buffer contains ammonium hydroxide (NH₄OH) and a salt of ammonia with a  strong acid like (HCl) which produces, ammonium chloride (NH₄Cl) mixture. The evaporation rate of ammonia is so high at room temperature thus on opening of the buffer solution the ammonia get evaporated very fast and the concentration of ammonia decreases which affect the pH of the buffer solution.

Thus the reason to put ammonia buffer in fume hood is explained.


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kozerog [31]
Insulation wraps because independent is the variable you are changing to affect the dependent variable (what you are measuring)
3 0
3 years ago
Please help ASAP<br> Will give brainliest
Leni [432]

Answer:  457.8

Explanation:

345.8*225/703.55

8 0
2 years ago
Determine the [h3o+] of a 0.210 m solution of formic acid.
Nataly [62]
When the Pka for formic acid = 3.77
and Pka = -㏒ Ka 
   3.77 = -㏒ Ka
∴Ka = 1.7x10^-4 

when Ka = [H+][HCOO-}/[HCOOH]

when we have Ka = 1.7x10^-4 &[HCOOH] = 0.21 m
so by substitution: by using ICE table value
1.7x10^-4 = X*X / (0.21-X)
(1.7x10^-4)*(0.21-X) = X^2      by solving this equation for X

∴X = 0.0059
∴[H+] = 0.0059
∴PH= -㏒ [H+]
       = -㏒ 0.0059
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3 0
3 years ago
The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by
maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use <em>Raoult’s Law</em> to calculate the vapour pressure:  

<em>p</em>₁ = χ₁<em>p</em>₁°  

where  

χ₁ = the mole fraction of the solvent  

<em>p</em>₁ and <em>p</em>₁° are the vapour pressures of the solution and of the pure solvent  

The formula for vapour pressure lowering Δ<em>p</em> is  

Δ<em>p</em> = <em>p</em>₁° - <em>p</em>₁  

Δ<em>p</em> = <em>p</em>₁° - χ₁<em>p</em>₁° = p₁°(1 – χ₁) = χ₂<em>p</em>₁°  

where χ₂ is the mole fraction of the solute.  

<em>Step 1</em>. Calculate the <em>mole fraction of glucose </em>

<em>n</em>₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu  

<em>n</em>₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O  

χ₂ = <em>n</em>₂/(<em>n</em>₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62  

<em>Step 2</em>. Calculate the <em>vapour pressure lowering</em>  

Δ<em>p</em> = χ₂<em>p</em>₁° = 0.018 62 × 23.8 torr = 0.4430 torr  

<em>Step 3</em>. Calculate the <em>vapour pressure</em>  

<em>p₁</em> = <em>p</em>₁° - Δ<em>p</em> = 23.8 torr – 0.4430 torr = 23.4 torr

3 0
3 years ago
What kind of reaction occurs when you mix aqueous solutions of barium sulfide and sulfuric acid?
Shkiper50 [21]
Precipitation and gas evolution
7 0
3 years ago
Read 2 more answers
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