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>
Answer:
P = 20atm
Explanation:
P1 = 10atm
T1 = 10K
P2 = ?
T2 = 20K
This question requires the use of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided its volume remains constant
Mathematically,
P = kT, k = P / T
P1 / T1 = P2 / T2 = P3 / T3=........=Pn / Tn
P1 / T1 = P2 / T2
P2 = (P1 × T2) / T1
P2 = (10 × 20) / 10
P2 = 20atm
The final pressure of the gas is 20atm
Glaciers are found in areas where the net accumulation of snow exceeds the melt, therefore they are usually found in areas of high elevation, cold temperatures, and high precipitation.
Geysers are found in areas of shallow crust with high heat flow, such as hot spots.
Solubility of a compound in water can be referred to as the amount of the compound that can be dissolved in 1 L of the solvent (water) at any given temperature. Solubility of a compound can be expressed in the units of g/L or mg/L.
Given that the solubility of calcium carbonate in water = 14 mg/L
We have to calculate the volume of water that can dissolve 11 g of calcium carbonate.
Converting 11 g calcium carbonate to mg:
Volume of water that would dissolve 11000 mg calcium carbonate
= 
=785.7 L
Rounding the volume 785.7 L to two significant figures, we get 790 L water.
Therefore, we would need 790 L water to completely dissolve 11 g of calcium carbonate.
If an unknown substance CANNOT be broken down into simpler substances, it is AN ELEMENT made of one kind of atom.
