Full Question:
A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?
2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)
Answer:
13.1 g K2CO3 required to neutralize spill
Explanation:
2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)
Number of moles = Volume * Molar Concentration
moles HBr= 0.42L x .45 M= 0.189 moles HBr
From the stoichiometry of the reaction;
1 mole of K2CO3 reacts with 2 moles of HBr
1 mole = 2 mole
x mole = 0.189
x = 0.189 / 2 = 0.0945 moles
Mass = Number of moles * Molar mass
Mass = 0.0945 * 138.21 = 13.1 g
Answer:
carboxylic acid
nitrile
sulfonic acid
aldehyde
Easter
ketone
acid halidle
alcohol
acid halide
thiol
amide
amine
Explanation:
its the answer but i can't explain
Answer:
Ammonia acts as an Arrhenius base because it increases the concentration of OH⁻ in aqueous solution.
Explanation:
The acid-base theory of Arrhenius explains that in aqueous solutions both acid and base dissociate, releasing ions in the solution. The acid release the ion H⁺ and some anion, and the base release the ion OH⁻ and some cation.
In water, the reaction of ammonia is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Because of that, ammonia is an Arrhenius base.
Answer:
The density of the sample is 11.8 g/mL.
Explanation:
We are given an unknown sample with a mass of 23.5 g and a volume of 2.00 mL and we want to determine its density.
Recall that density is given by:
Where ρ is the density, <em>m</em> is the mass, and <em>V</em> is the volume.
Substitute 23.5 g for <em>m</em> and 2.00 mL for <em>V</em> and evaluate:
Hence, the density of the sample is 11.8 g/mL.
Hi,
10,000,000sec = 16.53 weeks
Hope this helps.
r3t40
