Answer:
C.AgNO3 + NaCl → AgCl + NaNO3
Explanation:
2H2O=2H2+O2
37.4g H2O(1 mol/18.02)=2.07547 mol H2O
PV=nRT
(1.30)(V)=(2.07547)(.0821)(297)
Vwater=38.92898L
38.92898L (1 mol O2/2 mol H2O)=19.46449L O2 gas
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>
The relationship of radiation with distance obeys the inverse square law. Therefore, doubling the distance decrease the radiation by a factor of 4. The new count is 250.
1) Applying the same principle, the count decreases by a factor of 100. The new count is 10
2) An alpha particle is 4He2 and the Hydrogen can be represented as 1H1
14N7 + 4He2 - 1H1
= 17X8
Proton number 8 belongs to Oxygen. Therefore, the resultant nucleus is:
17O8
3) 185Au79 - 4He2
= 181Ir77
4) X - 4He2 = 234Th90
X = 238U92
5) Beta emission results in the same nucleon number but an increase in the proton number; therefore, the result is:
234Pa91