7.4x10^23 = molecules of silver nitrate sample
6.022x10^23 number of molecules per mole (Avogadro's number)
Divide molecules of AgNO3 by # of molecules per mol
7.4/6.022 = 1.229 mols AgNO3 (Sig Figs would put this at 1.3)
(I leave off the x10^23 because they both will divide out)
Use your periodic table to find the molar weight of silver nitrate.
107.868(Ag) + 14(N) + 3(16[O]) = 169.868g/mol AgNO3
Now multiply your moles of AgNO3 with your molar weight of AgNO3
1.229mol x 169.868g/mol = 208.767g AgNO3
Answer:
Graphics can sometimes convey more information in a brief amount of space than an author can explain in a paragraph.
S + O2 → SO2
<span>z / (32.0655 g S/mol) x (1 mol SO2 / 1 mol S) x (64.0638 g SO2/mol) = (1.9979 z) g SO2 </span>
<span>C + O2 → CO2 </span>
<span>(9.0-z) / (12.01078 g C/mol) x (1 mol CO2 / 1 mol C) x (44.00964 g CO2/mol) = (32.9776 - 3.66418 z) g CO2 </span>
<span>Add the two masses of SO2 and CO2 and set them equal to the amount given in the problem: </span>
<span>(1.9979 z) + (32.9776 - 3.66418 z) = 27.9 </span>
<span>Solve for z algebraically: </span>
<span>z = 3.0 g S</span>
a. t=0.553 s
b. vox(horizontal speed) = 3.62 m/s
<h3>Further explanation</h3>
Given
h = 1.5 m
x = 2 m
Required
a. time
b. vo=initial speed
Solution
Free fall motion
a. h = 1/2 gt²(vertical motion=h=voyt+1/2gt²⇒voy = 0)
t = √2h/g
t = √2.1.5/9.8
t=0.553 s
b. x=vox.t(horizontal motion)
vox=x/t
vox=2/0.553
vox=3.62 m/s
