Answer:
1.27 atm.
Explanation:
The following data were obtained from the question:
Pressure (P) in torr = 968 torr
Pressure (P) in atm =...?
Thus, we can convert from torr to atm by doing the following:
Recall:
760 torr = 1 atm
Therefore,
968 torr = 968/760 = 1.27 atm
Therefore, 968 torr is equivalent to 1.27 atm.
Answer:
Option a.
0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i
Explanation:
To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)
These are the colligative properties:
ΔP = P° . Xm . i → Lowering vapor pressure
ΔT = Kb . m . i → Boiling point elevation
ΔT = Kf . m . i → Freezing point depression
π = M . R . T → Osmotic pressure
Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.
CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:
CaCl₂ → Ca²⁺ + 2Cl⁻
We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3
KNO₃ → K⁺ + NO₃⁻
We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2
Option a, is the best.
Answer:
410.196 J/[kg*°C].
Explanation:
1) the equation of the energy is: E=c*m*(t₂-t₁), where E - energy (523 J), c - unknown specific heat of copper, m - mass of this copper [kg], t₂ - the final temperature, t₁ - initial temerature;
2) the specific heat of copper is:
![c=\frac{E}{m*(t_2-t_1)}; \ => \ c=\frac{523}{0.085*(45-30)}=\frac{523}{1.275}=410.196[\frac{J}{kg*C}].](https://tex.z-dn.net/?f=c%3D%5Cfrac%7BE%7D%7Bm%2A%28t_2-t_1%29%7D%3B%20%5C%20%3D%3E%20%5C%20c%3D%5Cfrac%7B523%7D%7B0.085%2A%2845-30%29%7D%3D%5Cfrac%7B523%7D%7B1.275%7D%3D410.196%5B%5Cfrac%7BJ%7D%7Bkg%2AC%7D%5D.)
Answer:
1) acetylide
2) enol
3) aldehydes
4) tautomers
5) alkynes
6) Hydroboration
7) Keto
8) methyl ketones
Explanation:
Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.
Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.
The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.
Answer:
(B) F⁻, HCOOH
Explanation:
(A) CH₄, HCOOH
(B) F⁻, HCOOH
(C) F⁻, CH₃-O-CH₃
The hydrogen bonds are formed when the hydrogen is found between two electronegative atoms such as oxygen (O), nitrogen (N) or florine (F).
O····H-O, F····H-O, O····H-N
(A) CH₄, HCOOH
- here methane CH₄ is not capable to form hydrogen bond with water
- formic acid HCOOH can form hydrogen bonds with water
H-C(=O)-O-H····OH₂
(B) F⁻, HCOOH
-both floride (F⁻) and formic acid can form hydrogen bonds with water
F····OH₂
H-C(=O)-O-H····OH₂
(C) F⁻, CH₃-O-CH₃
- dimethyl-ether CH₃-O-CH₃ is not capable to form hydrogen bond with water
- floride (F⁻) can form hydrogen bonds with water
F····OH₂
