Answer:
14 mL
Explanation:
To prepare a solution by a concentrated solution, we must use the equation:
C1xV1 = C2xV2, where <em>C</em> is the concentration, <em>V</em> is the volume, 1 is the initial solution and 2 the final solution.
The final solution must have 2 mL and a concentration of 350 pg/mL, and the initial solution has a concentration of 50 pg/mL.
Then:
50xV1 = 350x2
50xV1 = 700
V1 = 700/50
V1 = 14 mL
Answer:
The compound which are boh soluble in water and hexane is
B. Ethanol and 1-propanol
Explanation:
The compounds ethanol and 1-propanol are soluble in both hexane and water.
It is soluble in water as both consists of polar end due to hydrogen bonding present in the -OH functional group.
and both are soluble in hexane as they contain a non polar end and the alliphatic hydrocarbon chain in them.
The solubility of alcohols varies in increasing order as the hydrocarbon chain increases. And becaue of this it becomes more non polar.
Non polar properties decreases for branched molecules.
so, the correct option is ethanol and 1-propanol.
Answer:
Gas
Explanation:
This is kind of like the process of evaporating
Answer:
0.861 L
Explanation:
We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:
(P₁V₁)/T₁ = (P₂V₂)/T₂
Convert the temperatures to degrees Kelvin.
25.0°C -> 298 K, 100.0°C -> 373 K
Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:
(165(2.5))/298 = (600(V₂))/373
V₂ = (165(2.5)(373))/(298(600))
V₂ = 0.861 L
What amount of heat absorbs 50 g of steel (ce = 0.115 cal / g. ° C) that
does its temperature vary by 25 ° C?
Answer:
143.75cal
Explanation:
Given parameters:
Mass of steel = 50g
Specific heat capacity of the steel = 0.115cal/g°C
Temperature = 25°C
Unknown:
Amount of heat = ?
Solution:
The amount of heat to cause this temperature change is dependent on mass and specific heat capacity of the substance.
Amount of heat = m C (ΔT)
m is the mass
c is the specific heat capacity
ΔT is the temperature change
Now insert the parameters and solve;
Amount of heat = 50 x 0.115 x 25
Amount of heat = 143.75cal
