Answer:
464.29mL de solvente y 35.71mL de ácido nítrico concentrado deben agregarse.
Explanation:
El ácido nítrico concentrado viene al 70% v/v por temas de estabilidad. El volumen de ácido nítrico que se debe agregar si se quieren hace 500mL al 5% de HNO3 es:
500mL * (5mL / 100mL) = 25mL de ácido nítrico se deben agregar.
Como el ácido nítrico está al 70%:
25mL ácido nítrico * (100mL / 70mL ácido nítrico) = 35.71mL de ácido nítrico concentrado deben agregarse.
Y el volumen de solvente debe ser:
500mL - 35.71mL = 464.29mL
Answer:
Sugar, sodium chloride, and hydrophilic proteins
Answer:
Explanation:
Data:
p₁ = 694.9 mmHg; V₁ = 3.463 L
p₂ = ?; V₂ = 5.887 L
Calculation:
Explanation:
The endoplasmic reticulum consists of a network of a tube-like passageway through which proteins from the ribosomes are able to be moved within a cell as the road system allows for movement throughout the city.
We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams
