What question? Lhh this is hilarious.
<span>1.11 g/cm³
Hope this helps! </span>
Answer:
0.92^n
Explanation:
Given that :
Initial amount of vinegar = 1 Litre
Number of litres removed repeatedly = 0.08 Litre
Since the amount removed each time is constant, then ;
Initial % = 100% = 100/100 = 1
. Using the relation :
Amount of vinegar in mixture :
Initial * (1 - amount removed / initial amount)^n
n = number of times repeated
1 * (1 - 0.08/1)^n
1 * (1 - 0.08)^n
1 * 0.92^n
Hence,
For nth removal,
Concentration will be :
0.92^n ; for n ≥ 1
Answer:
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
Explanation:
The pH of the solution = 13.00
pH + pOH = 14
pOH = 14 - pH = 14 - 13.00 = 1.00
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![1.00=-\log[OH^-]](https://tex.z-dn.net/?f=1.00%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-1.00} M=0.100 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-1.00%7D%20M%3D0.100%20M)

![[KOH]=[OH^-]=[K^+]=0.100 M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5BK%5E%2B%5D%3D0.100%20M)
Molariy of the KOH = 0.100 M
Volume of the KOH solution = 800 mL= 0.800 L
1 mL = 0.001 L
Moles of KOH = n


n = 0.0800 mol
Mass of 0.0800 moles of KOH :
0.0800 mol × 56 g/mol = 4.48 g
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
Answer:
0.0303 Liters
Explanation:
Given:
Mass of the potassium hydrogen phosphate = 0.2352
Molarity of the HNO₃ Solution = 0.08892 M
Now,
From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃
The number of moles of 0.2352 g of potassium hydrogen phosphate
= Mass / Molar mass
also,
Molar mass of potassium hydrogen phosphate
= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol
Number of moles = 0.2352 / 174.15 = 0.00135 moles
thus,
The number of moles of HNO₃ required for 0.00135 moles
= 2 × 0.00135 mol of HNO₃
= 0.0027 mol of HNO₃
Now,
Molarity = Number of Moles / Volume
thus,
for 0.0027 mol of HNO₃, we have
0.08892 = 0.0027 / Volume
or
Volume = 0.0303 Liters