Answer:
An emulsion is formed.
Explanation:
An association (emulsion) of two liquids is formed, in this case oil and vinegar, which when stirred, said mixture will separate.
Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).
We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.
Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.
(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).
Answer:
Sucrose is a disaccharide composed of alpha D gluose and beta D fructose linked together by beta 2,alpha1 glycosidic linkage.
Explanation:
The specificity of glycosidic linkage very much essential to choose the substrate for the synthesis of specific disaccharide.
For example sucrose contain beta 2,alpha1 glycosidic linkage that means the hydroxyl group of anomeric carbon of one monosaccharide(fructose) should remain in beta conformation and the hydroxyl group of other monosaccharide(glucose) should remain in alpha conformation.
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.
Here, the three different notation of the p-orbital in different sub-level have to generate
The value of azimuthal quantum number (l) for -p orbital is 1. We know that the magnetic quantum number 
depends upon the value of l, which are -l to +l.
Thus for p-orbital the possible magnetic quantum numbers are- -1, 0, +1. So there will be three orbitals for p orbitals, which are designated as 
, 
and 
in space.
The three p-orbital can be distinguish by the quantum numbers as-
For 2p orbitals (principal quantum number is 2)
1) n = 2, l = 1, m = -1
2) n = 2, l = 1, m = 0
3) n = 2, l = 1, m = +1
Thus the notation of different p-orbitals in the sub level are determined.
