Answer:
Explanation:
Whenever you see molar masses in gas law questions, more often than not density will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation PV=nRT to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:M=dRTPM = molar mass (g/mol)d = density (g/L)R = Ideal Gas Constant (≈0.0821atm⋅Lmol⋅K) T = Temperature (In Kelvin) P = Pressure (atm)As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the ideal gas law in which the identity of the gas plays a role in your calculations. Just something to take note of. Back to the problem: Now, looking back at what we're given, we will need to make some unit conversions to ensure everything matches the dimensions required by the equation:T=35oC+273.15= 308.15 KV=300mL⋅1000mL1L= 0.300 LP=789mmHg⋅1atm760mmHg= 1.038 atmSo, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:d=0.622g0.300L= 2.073 g/LNow, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.M=dRTP=(2.073)(0.0821)(308.15)1.038= 51 g/molRounded to 2 significant figuresNow if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol). Hope that helped :)
The conclusion that would support the students prediction is B) Plant "A" grows taller than Plant "B".
If the student thinks that adding fertilizer to the plant would help it grow then answer B) would make the most sence.
Answer:
10 Litre
Explanation:
Given that ::
v1 = 25L ; n1 = 1.5 mole ; v2 =? ; n2 = (1.5-0.9) = 0.6 mole
Using the relation :
(n2 * v1) / n1 = (n2 * v2) / n2
v2 = (n2 * v1) / n1
v2 = (0.6 mole * 25 Litre) / 1.5 mole
v2 = 15 / 1.5 litre
v2 = 10 Litre
It has a higher boiling point and a lower freezing point
Answer:
0.471 mol/L
Explanation:
First, we'll begin by by calculating the number of mole of KMnO4 in 26g of KMnO4.
This is illustrated below:
Molar Mass of KMnO4 = 39 + 55 + (16x4) = 39 + 55 + 64 = 158g/mol
Mass of KMnO4 from the question = 26g
Mole of KMnO4 =?
Number of mole = Mass/Molar Mass
Mole of KMnO4 = 26/158 = 0.165mole
Now we can obtain the concentration of KMnO4 in mol/L as follow:
Volume of the solution = 350mL = 350/1000 = 0.35L
Mole of KMnO4 = 0.165mole
Conc. In mol/L = mole of solute(KMnO4)/volume of solution
Conc. In mol/L = 0.165mol/0.35
conc. in mol/L = 0.471mol/L
