Answer:
Specific heat is defined as the amount of heat needed to raise the temperature by one degree celsius. Therefore, in 1 kg there are 100 grams so, 10 grams equal 0.01 Kg. Thus, calculate the specific heat value as follows. Thus, we can conclude that specific heat of the given metal is 0.5 .
Explanation:
Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.
Answer:
(1) Bromination, (2) E2 elimination and (3) epoxidation
Explanation:
- In the first step, -OH group in cyclopentanol is replaced by more facile leaving group Br by treating cyclopentanol with
- In the second step, E2 elimination in presence of strong base e.g. NaOEt/EtOH produce cyclopentene
- In the third step, treatment of cyclopentene with mCPBA produces 1,2-epoxycyclopentane
- Full reaction scheme has been shown below
Kepler’s third law exhibits the relationships between the distance of a planet from the sun and the period of its revolution. Kepler’s third law is also sometimes referred to as the law of harmonies.
Kepler’s third law compares the orbital period and the radius of an orbit of a planet to the distance of the planet to the sun. It states mathematically that the more distant a planet is from the sun the greater its orbital period will be. The period of revolution of a planet is measured in days, weeks, months or years. For example, Earth’s period of revolution is 365 days.
Answer:
XY4Z2 ----- square planar
XY5Z ------- square pyramidal
XY2Z----- bent < 120°
XY2Z3 ----- Linear
XY4Z ---- see saw
XY2Z2 ----- bent <109°
XY3Z2 ----- T shaped
XY3Z ----- Trigonal pyramidal
Explanation:
The valence shell electron pair repulsion theory ( VSEPR) gives the description of molecular geometry based on the relative number of electron pairs present in the molecule.
However, electron pairs repel each other, the repulsion between two lone pairs is greater than the repulsion between a lone pair and a bond pair which is also greater than the repulsion between two lone pairs.
The presence of lone pairs distort the bond angle and molecular geometry from the expected geometry based on VSEPR theory. Hence, in the presence of lone pairs of electron, the observed molecular geometry may be different from that predicted on the basis of the VSEPR theory, the bond angles also differ slightly or widely depending on the number of lone pairs present.
All the molecules in the question possess lone pairs, the number of electron pairs do not correspond to the observed molecular shape or geometry due to lone pair repulsion. Usually, the molecular geometry deals more with the arrangement of bonded atoms in the molecule.
