Azane
And it's Molar Mass : 17.031 g/mol
Answer:
The correct answer is 190.5 mL of 1.00 M KH₂PO₄
Explanation:
A phosphate buffer is composed by phosphate acid (KH₂PO₄) and its conjugated base (K₂HPO₄). To obtain the relation between the concentrations of base and acid to add, we use Henderson-Hasselbach equation:
pH= pKa + log 
We have: pH= 6.97 and pKa= 7.21. So, we replace the values in the equation:
6.97= 7.21 + log
6.97-7.21= log
-0.24= log
=
0.575 =
=
It means that you have to mix a volume 0.575 times of conjugated base and 1 volume of acid. If we assume a total buffer concentration of 1 M, we have:
base + acid = 1
base= 1 - acid
We replace in the previous equation:
0.575= 
0.575 acid= 1 - acid
0.575 acid + 1 acid= 1
1.575 acid = 1
acid= 1/1,575
acid= 0.635
base= 1 - acid = 1 - 0.635 = 0.365
For a total volume of 300 ml, the volumes of both acid and base will be:
300 ml x 0.635 M = 190.5 ml of acid (KH₂PO₄)
300 ml x 0.365 M= 109.5 ml of base (K₂HPO₄)
We can corroborate our calculations as follows:
190.5 ml + 109.5 ml = 300 ml
109.5 ml / 190.5 ml = 0.575
Ammonium chloride is salt, and when we add solid salt into water, it’s dissolved and nothing happens. The only thing happens when any salt dissolved in water, is becoming its ions and improve its conductivity.
NH4Cl(s) + H2O(l) → NH4+(aq) + Cl-(aq) + H2O(l)
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<span>Theory of Matter state that matter is composed of?
</span><span>
Answer : Large number of small particles—individual atoms or molecules.</span>
Answer:
68.6 °C
Explanation:
From conservation of energy, the heat lost by acetone, Q = heat gained by aluminum, Q'
Q = Q'
Q = mL where Q = latent heat of vaporization of acetone, m = mass of acetone = 3.33 g and L = specific latent heat of vaporization of acetone = 518 J/g
Q' = m'c(θ₂ - θ₁) where m' = mass of aluminum = 44.0 g, c = specific heat capacity of aluminum = 0.9 J/g°C, θ₁ = initial temperature of aluminum = 25°C and θ₂ = final temperature of aluminum = unknown
So, mL = m'c(θ₂ - θ₁)
θ₂ - θ₁ = mL/m'c
θ₂ = mL/m'c + θ₁
substituting the values of the variables into the equation, we have
θ₂ = 3.33 g × 518 J/g/(44.0 g × 0.9 J/g°C) + 25 °C
θ₂ = 1724.94 J/(39.6 J/°C) + 25 °C
θ₂ = 43.56 °C + 25 °C
θ₂ = 68.56 °C
θ₂ ≅ 68.6 °C
So, the final temperature (in °C) of the metal block is 68.6 °C.
