The molecular formula of calcium oxide - CaO
The molar mass of CaO - 40 + 16 = 56 g/mol
Which means that 1 mol weighs 56 g
Therefore 56 g of CaO is - 1 mol
Then 89.23 g is equivalent to - 1/56 x 89.23 = 1.6 mol of CaO
<span>NaCl
First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound
Sodium = 22.989769
Chlorine = 35.453
Silver = 107.8682
Nitrogen = 14.0067
Oxygen = 15.999
Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results
For NaCl
22.989769 + 35.453 = 58.44277
For AgNO3
107.8682 + 14.0067 + 3 * 15.999 = 169.8719
Now calculate how many moles of each substance by dividing the total mass by the molar mass
For NaCl
4.00 g / 58.44277 g/mol = 0.068443 mol
For AgNO3
10.00 g / 169.8719 g/mol = 0.058868
Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>
Energy cannot be destroyed or created, but energy could be transformed or transferred. For example a skiier skiing from the mouth can have potential energy transferred into kinetic energy.
The gas will occupy 120 L at STP.
We can use the <em>Combined Gas Laws</em>
.
In 1982, chemists redefined <em>STP</em> as <em>1 bar</em> and 0 °C (273.15 K).
<em>p</em>_1 = 5100 mmHg × (1 atm/760 mmHg) × (1.013 bar/1 atm) = 6.798 bar
<em>T</em>_1 = (29 + 273.15) K = 302.15 K
<em>p</em>_1<em>V</em>_1/<em>T</em>_1 = <em>p</em>_2<em>V</em>_2/<em>T</em>_2
<em>V</em>_2 = <em>V</em>_1 × <em>p</em>_1/<em>p</em>_2 × <em>T</em>_2/<em>T</em>_1
= 20.1 L × (6.798 bar/1 bar) × (273.15 K/302.15 K) = 120 L
We are given with
V = <span>2.394 x 102 mL
P = </span><span>7.20 x 102 mm of Hg
T = </span><span>78 oC
We are asked to determine the mass of the sample in milligrams
Using the ideal gas law
n = PV / RT
n = </span>7.20 x 102 mm of Hg (2.394 x 102 mL) / R (78 + 273)
Use the appropriate value for R or just convert the values to SI and use R = 8.314
Then, solve for the mass of the sample by using the MW Oxygen which is 16 g/mol
