Answer:
(a) 77.9 g/mol
(b) 3.18 g / L
Explanation:
<u>(a)</u> We need to use the ideal gas law, which states: PV = nRT, where P is the pressure, V is the volume, n is the moles, R is the gas constant, and T is the temperature in Kelvins.
Notice that we don't have moles; we instead have the mass. Remember, though that moles can be written as m/M, where m is the mass and M is the molar mass. So, we can replace n in the equation with m/M, or 21.3/M. The components we now have are:
- P: 0.880 atm
- V: 7.73 Litres
- n: m/M = 21.3 g / M
- R: 0.08206
- T: 30.00°C + 273 = 303 K
Plug these in:
PV = nRT
(0.880)(7.73) = (21.3/M)(0.08206)(303)
Solve for M:
M = 77.9 g/mol
<u>(b)</u> The equation for the molar mass is actually:
M = (dRT)/P, where d is the density
We have all the components except d, so plug them in:
77.9 = (d * 0.08206 * 298) / 1
Solve for d:
d = 3.18 g / L
Awnser your loooking fo its got to be aluminium
Explanation:i took the dang subgect
Answer:
AgNO3 + NaOH = AgOH + NaNO3.
Explanation:
<em><u>Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.</u></em>
<em><u>Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.</u></em>
To begin this, you'll first note down the valencies of atoms you're sure about.
K is always 1+
And O is always 2-, except for in peroxide where it is 1-.
So now you must remember that a compound is always neutral. Your net charge has to be 0.
K2Cr2O7.
K2 (1+ x 2) = 2+
Cr (2x) = 2x
O7 (2- ×7) = 14-
+2 +2x + (-14) = 0
•Simple algebra.
2x -12 = 0
2x = 12
x = +6.
Each Cr atom has an oxidation number of +6.
