Answer:
Gaseous nitrogen has unique chemical and physical properties that make it suitable for use in food processing. Nitrogen is inert which means it will not react with prepared food materials, which can alter their aromas or flavors. Also, gaseous nitrogen will effectively displace oxygen minimizing oxidation and the growth of microorganisms that cause foods to lose their freshness and deteriorate faster.
Explanation:
Source: https://www.generon.com/using-nitrogen-gas-in-food-packaging/
Answer:
Conditioning two or three times will insure that the concentration of titrant is not changed by a stray drop of water.
Explanation:
"Check the tip of the buret for an air bubble. To remove an air bubble, whack the side of the buret tip while solution is flowing".
Answer:
<em>The correct option is A) Arrhenius</em>
Explanation:
According to the Arrhenius concept of acids and bases, an acid must produce H+ ions when it is present in a solution and the base must produce OH- ions when placed in a solution.
Ammonia does not contain OH- ions of its own when dissolved in water.
The reaction of ammonia dissolving is water can be written as:
NH3 + H2O ⇌ NH4+ + OH−
As we can see from the equation, ammonia does form OH- ions but it does not have OH- ions on its own.
Hence, according to the Arrhenius concept, NH3 is not a base.
Answer:
See explanation below
Explanation:
In order to calculate this, we need to use the following expression to get the concentration of the base:
MaVa = MbVb (1)
We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:
Atomic weights of the elements to be used:
K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol
MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol
Now, let's calculate the mole of KHP:
moles = 0.5053 / 204.2189 = 0.00247 moles
With the moles, we also know that:
n = M*V (2)
Replacing in (1):
n = MbVb
Now, solving for Mb:
Mb = n/Vb (3)
Finally, replacing the data:
Mb = 0.00247 / (13.4473/1000)
Mb = 0.184 M
This would be the concentration of NaOH
