Answer:
0.234 M
Explanation:
C- 12.009 x 7
H- 1.001 x 5
N- 14.006
O- 16 x 3
S- 32.059
___________+
183.133 g/mol
= 0.234 M Cancel out the grams mol/L equals molarity. Lowest significant figure is 3
Answer:
177.1 L
Explanation:
The excersise can be solved, by the Ideal Gases Law.
P . V = n . R . T
In first step we need to determine the moles of gas:
We convert T° from, C° to K → 20°C + 273 = 293K
We convert P from mmHg to atm → 760 mmHg = 1atm
1Dm³ = 1L → 190L
We replace: 190 L . 1 atm = n . 0.082 . 293K
(190L.atm) / 0.082 . 293K = 7.91 moles.
We replace equation at STP conditions (1 atm and 273K)
V = (n . R .T) / P
V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L
We can also make a rule of three:
At STP conditions 1 mol of gas occupies 22.4L
Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Answer:
Lines of latitude run from east to west.
Explanation:
Answer:
Explanation:
The first one is CrO. The Chromium has the same charge as the oxygen so mol numbers are dropped.
The Second one is CrO2 The two oxygens have a charge of 2(-2) = -4. To balance this, the Chromium must have a charge of +4 Cr(Iv)O2
The third one is can be set up like this
Cr + 3(-2) = 0
Cr - 6 = 0
Cr = 6
Therefore the formula is Cr(vi)O3
The last one is a bit tricky. Follow this carefully. There are 2 Crs and 3Os.
The formula looks like this
2Cr + 3(-2) = 0
2Cr - 6 = 0
2Cr = 6
Cr = 3
The formula is Cr(iii)2 O3
