Answer:GLACIERS ARE MADE UP OF FALLEN SNOW OVER MILLION YEARS, IN MOST OF POLAR REGIONS IT TURNS TO ICE AND WILL FLOW DOWNWARDS AND OUTWARDS UNDER ITS OWN PRESSURE
AND THE LARGEST GLACIER IN THE WORLD IS 60 MILES WIDE AND AROUND 270 MILES LONG.
IT COVERS 10% OF EARTH'S TOTAL LAND
Explanation:
Answer:
#1: 0.00144 mmolHCl/mg Sample
#2: 0.00155 mmolHCl/mg Sample
#3: 0.00153 mmolHCl/mg Sample
Explanation:
A antiacid (weak base) will react with the HCl thus:
Antiacid + HCl → Water + Salt.
In the titration of antiacid, the strong acid (HCl) is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.
Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:
Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl
Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl
Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl
The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.
Thus, mmoles HCl /mg OF SAMPLE<em> </em>for each trial is:
#1: 2.178mmol / 1515mg
#2: 2.253mmol / 1452mg
#3: 2.219mmol / 1443mg
<h3>#1: 0.00144 mmolHCl/mg Sample</h3><h3>#2: 0.00155 mmolHCl/mg Sample</h3><h3>#3: 0.00153 mmolHCl/mg Sample</h3>
Explanation:
The problem basically wants you to find a way to convert between the number of atoms present in the sample and the number of moles they are equivalent to.
To convert between atoms and moles we use something called Avogadro's constant, which basically acts as the definition of a mole.
More specifically, in order to have one mole of an element you need
6.022 x 10^23 atoms of that element. You can thus use this number as a conversion factor to take you from atoms to moles or vice versa.
In your case, you will have
3.90 x 10^ 26 atoms Zn x 1 mole Zn ( Avogrado’s constant) / 6.022 x 10^23 atoms Zn
= 6.5 x 10^8 is the answer
Answer:
Explanation:
1. Calculate the initial moles of acid and base
2. Calculate the moles remaining after the reaction
OH⁻ + H₃O⁺ ⟶ 2H₂O
I/mol: 0.0053 0.005 00
C/mol: -0.00500 -0.005 00
E/mol: 0.0003 0
We have an excess of 0.0003 mol of base.
3. Calculate the concentration of OH⁻
Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L
![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0003%20mol%7D%7D%7B%5Ctext%7B0.078%20L%7D%7D%20%3D%20%5Ctextbf%7B0.0038%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20final%20concentration%20of%20OH%24%5E%7B-%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.0038%20mol%2FL%7D%7D%24%7D)
