Answer:
<em>In a physical change the appearance or form of the matter changes but the kind of matter in the substance does not. However in a chemical change, the kind of matter changes and at least one new substance with new properties is formed.</em>
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I HOPE THIS WILL HELP YOU IF NOT THEN SORRY HAVE A GREAT DAY:)</h3>
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>
Answer:
110V
Explanation:
Given parameters:
Current in the lightbulb = 0.5A
Resistance = 220Ω
Unknown:
Voltage = ?
Solution:
To solve this problem, we apply the equation derived from ohm's law;
V = IR
So,
V = 0.5 x 220 = 110V
Dalton postulated that atoms of the same elements had the same mass while atoms of different elements had different masses.
The discovery of isotopes revealed that atoms of the same element did not necessarily have same masses, they could have different masses.
Therefore, the modern periodic table that we use nowadays contains the average atomic mass of all the isotopes found for a given element.