All atoms consist of a nucleus with protons and neutrons that is surrounded by a sea of electrons. If Uranium-235 is bombarded with a neutron then the atom's atomic number will remain the same but its mass will increase by 1 unit. Thus, it will temporarily become Uranium-236.
212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.
So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then, 
So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
B) The term "inert" was dropped because it no longer described all the group 8A elements.
Explanation:
Inert elements in chemistry simply refers to elements that are chemically inactive and are not expected to form any compounds. this is the general belief for the group 8 elements as they all have complete duplet/octet configurations (and ideally, they ought to be very stable with no tendency to form compounds by participating in the loss and gain of electrons). However the discovery of compounds like xenon tetrafluoride (XeF4) proved this to be wrong.
Again, the reason the term - inert gses was droppedis beacause this term is not strictly accurate because several of them do take part in chemical reactions.
After dropping the term - Inert gases, they are now referred to as noble gases.
Answer:
15.70mg would remain
Explanation:
Partition coefficient is used to extract or purify a solute from a solvent selectively to avoid interference from other substances. For the problem, formula is:
Kp = Concentration 9-fluorenone in ether / Concentration of solute in H₂O
After the solute, 9-fluorenone, is extracted with water, the mass that remains in ether is:
(19mg - X)
<em>Where X is the mass that now is in the aqueous phase</em>
Replacing in Kp formula:
9.5 = (19mg - X) / 5mL / (X /10mL)
0.95X = 19mg - X / 5mL
4.75X = 19 - X
5.75X = 19
X = 19 / 5.75
X = 3.30mg
That means 9-fluorenone that remain in the ether layer is:
19mg - 3.30mg =
<h3>15.70mg would remain</h3>
Answer:
8740 joules are required to convert 20 grams of ice to liquid water.
Explanation:
The amount of heat required (
), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:
![Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}]](https://tex.z-dn.net/?f=Q%20%3D%20m%5Ccdot%20%5Bc%5Ccdot%20%28T_%7Bf%7D-T_%7Bo%7D%29%2BL_%7Bf%7D%5D)
(1)
Where:
- Mass, measured in grams.
- Specific heat of ice, measured in joules per gram-degree Celsius.
, 
- Temperature, measured in degrees Celsius.
- Latent heat of fussion, measured in joules per gram.
If we know that 
, 
, 
, 
and 
, then the amount of heat is:
![Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}](https://tex.z-dn.net/?f=Q%20%3D%20%2820%5C%2Cg%29%5Ccdot%20%5Cleft%5C%7B%5Cleft%282.06%5C%2C%5Cfrac%7BJ%7D%7Bg%5Ccdot%20%5E%7B%5Ccirc%7DC%7D%20%5Cright%29%5Ccdot%20%5B0%5C%2C%5E%7B%5Ccirc%7DC-%28-50%5C%2C%5E%7B%5Ccirc%7DC%29%5D%2B334%5C%2C%5Cfrac%7BJ%7D%7Bg%7D%20%5Cright%5C%7D)
8740 joules are required to convert 20 grams of ice to liquid water.
