Answer:
13.5g of AgNO3 will be needed
Explanation:
Silver nitrate, AgNO3 contains 1 mole of silver, Ag, per mole of nitrate. To solve this problem we need to convert the mass of Ag to moles. Thee moles = Moles of AgNO3 we need. With the molar mass of AgNO3 we can find the needed mass:
<em>Moles Ag-Molar mass: 107.8682g/mol-</em>
8.6g * (1mol / 107.8682g) = 0.0797 moles Ag = Moles AgNO3
<em>Mass AgNO3 -Molar mass: 169.87g/mol-</em>
0.0797 moles Ag * (169.87g/mol) =
<h3>13.5g of AgNO3 will be needed</h3>
Answer:
Gas in motion : Vaporization
Examples of fluid flow :
- A river flowing down a mountain
- Air passing over a bird's wing
- Blood moving through a circulatory system
- Fuel moving through an engine.
Explanation:
Answer:
The answer to the question is
50 % of the original amount of potassium 40 will be left after one half life or 1.25 billion years
Explanation:
To solve the question we note that the half life is the time for half of the quantity of substance that undergoes radioactive decay to disintegrate, thus
we have
half life of potassium 40 K₄₀ = 1.25 billion years
To support the believe tht the rock was formed 1.25 billion years ago we have

After 1.25 billion years we have
=
=0.5 of
will be left or 50 % of the original amount of potassium 40 will be left
Answer:
The answer to your question is:
Vol of NO2 = 11.19 L
Vol of O2 = 2.8 L
Explanation:
Data
N2O5 = 56 g
STP T = 0°C = 273°K
P = 1 atm
MW N2O5 = 216 g
Gases law = PV = nRT
Process
216 g of N2O5 ---------------- 1 mol
54 g ----------------- x
x = (54 x 1) / 216
x = 0.25 mol of N2O5
2 mol of N2O5 ----------------- 4 mol of NO2
0.25 mol ------------------ x
x = (0.25 x 4) / 2 = 0.5 mol of NO2
V = nRT/P
V = (0.5)(0.082)(273) / 1 = 11.19 L
2 mol of N2O5 ----------------- 1 O2
0.25 N2O5 ---------------------- x
x = (0.25 x 1) / 2 = 0.125 mol
Vol = (0.125)((0.082)(273) / 1 = 2.8 L