If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
Answer:
When you are exercising, your muscles need extra oxygen—some three times as much as resting muscles. This need means that your heart starts pumping faster, which makes for a quicker pulse. Meanwhile, your lungs are also taking in more air, hence the harder breathing.
Explanation:
Answer:
Gly-Ala
(Dipeptide).
Explanation:
A condensation reaction is a reaction in which two molecules combine to form a single molecule with the loss of a small molecule e.g water.
When two amino acids, such as alanine and glycine react, they yield a mixture of dipeptides (via dipeptide bond) and water. That is, The −OH from the carboxyl group (-COOH) of one amino acid combines with a hydrogen atom from the amine group (-NH2) of the other amino acid to produce water.
Answer:
1.7x10²³ Fluorine atoms are present
Explanation:
To solve this question we must know that, to find the amount of atoms of a substance from its mass, we have, as first, to convert this mass to moles using molar mass of the substance. With the moles we can find the amount of molecules using Avogadro's number. And with the formula we can know the atoms presents:
<em>Moles Fluorine (F₂, Molar mass: 37.997g/mol)</em>
5.3g * (1mol / 37.997g) = 0.139 moles F₂
<em>Molecules F₂:</em>
0.139 moles F₂ * (6.022x10²³molecules / mol) = 8.4x10²² molecules F₂
As 1 molecule of F₂ contains 2 atoms of Fluorine:
8.4x10²² molecules F₂ * 2 =
<h3>1.7x10²³ Fluorine atoms are present</h3>
Oh wow there could be so many but if you take any ionic compound that is aqueous, I’m sure it would work? Aqueous is a solid that is dissolved in water by the way. So I mean, that would technically create a liquid solution.
Sorry I couldn’t be a better help.. I probably didn’t explain that very well either but I hope that kinda gives you an idea of an option. :\
