Answer:
A concentrated acid is an acid which is in either pure form or has a high concentration. Laboratory type sulfuric acid (about 98% by weight) is a concentrated (and strong) acid. A dilute acid is that in which the concentration of the water mixed in the acid is higher than the concentration of the acid itself.
Explanation:
Concentrated acid - Those acids which are pure or have very high concentration in water are called as concentrated acids. For example concentrated Hydrochloric acid (HCl) and concentrated Sulphuric acid are examples of concentrated acids.
The answer would be option D. can only be seperated by chemical means
Answer:
N2
Explanation:
We use the ideal gas equation to calculate the number of moles of the diatomic gas. Then from the number of moles we can get
Given:
P = 2atm
1atm = 101,325pa
2atm = 202,650pa
T = 27 degrees Celsius = 27 + 273.15 = 300.15K
V = 2.2L
R = molar gas constant = 8314.46 L.Pa/molK
PV = nRT
Rearranging n = PV/RT
Substituting these values will yield:
n = (202,650 * 2.2)/(8314.46* 300.15)
n = 0.18 moles
To get the molar mass, we simply divide the mass by the number of moles.
5.1/0.18 = 28.5g/mol
This is the closest to the molar mass of diatomic nitrogen N2.
Hence, the gas is nitrogen gas
Answer:
a visible record (such as a series of colored bands, or a graph) showing the result of separation of the components of a mixture by chromatography.
Explanation:
Chromatography is a laboratory technique for the separation of a mixture. The mixture is dissolved in a fluid called the mobile phase, which carries it through a system on which is fixed a material called the stationary phase.
Answer:
5a. 13.5 moles of F₂
5b. 3.33 moles of Ar.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
3F₂ + 2Ar —> 2ArF₃
From the balanced equation above,
3 moles of F₂ reacted with 2 moles of Ar to produce 2 moles of ArF₃.
5a. Determination of the number of mole of fluorine, F₂, needed to produce 9 moles of ArF₃.
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of ArF₃.
Therefore, Xmol of F₂ will react to produce 9 moles of ArF₃ i.e
Xmol of F₂ = (3 × 9)/2
Xmol of F₂ = 13.5 moles
Thus, 13.5 moles of F₂ is needed.
5b. Determination of the number of mole of Ar needed to react with 5 moles F₂.
From the balanced equation above,
3 moles of F₂ reacted with 2 moles of Ar.
Therefore, 5 moles of F₂ will react with = (5 × 2)/3 = 3.33 moles of Ar.
Thus, 3.33 moles of Ar was consumed.
