Acid A, assuming the two acids have the same pH. The M stands for molarity which is how concentrated a substance is (basically the higher the molarity the more concentrated the acid is). However, pH refers to how acidic a substance is. If the two acids have different levels of acidity, the answer may be different.
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:
Total energy required to raise the temperature of 425 g of tin from 298.15 K to 505.05 K and to melt the tin at 505.05 K is 45.249 kiloJoules.
Explanation:
Mass of the tin ,m= 425 g
Heat capacity of the tin ,c= 0.227 J/g K
Initial temperature of the tin ,
= 25.0 °C = 298.15 K
Final temperature of the tin,
= 231.9 °C = 505.05 K
Let the heat required to change the temperature of tin from 298.15 K to 505.05 K be Q.


Heat required to melt tin at 505.05 K be Q'
The heat of fusion of tin metal =

Total energy required to raise the temperature of 425 g of tin from 298.15 K to 505.05 K and to melt the tin at 505.05 K is:
= Q+Q' = 19.961 kJ + 25.288 kJ = 45.249 kJ
Hydrogen, helium and oxygen
Answer:
The volume of nitrogen oxide formed is 35.6L
Explanation:
The reaction of nitric acid with copper is:
Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)
Moles of copper are:

Moles of nitric acid are:

As 1 mol of Cu reacts with 4 moles of HNO₃:
0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.
Moles of NO₂ produced are:
0.697 mol Cu × (2mol NO₂ / 1mol Cu) = <em>1.394 moles of NO₂</em>
Using PV = nRT
<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>
Thus, volume is:
V = nRT / P
V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm
V = 35.6L
<em>The volume of nitrogen oxide formed is 35.6L</em>