Answer:
c = 4016.64 j/g.°C
Explanation:
Given data:
Mass of substance = 2.50 g
Calories release = 12 cal (12 ×4184 = 50208 j)
Initial temperature = 25°C
Final temperature = 20°C
Specific heat of substance = ?
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Solution:
Q = m.c. ΔT
ΔT = T2 - T1
ΔT = 20°C - 25°C
ΔT = -5°C
50208 j = 2.50 g . c. -5°C
50208 j = -12.5 g.°C .c
50208 j /-12.5 g.°C = c
c = 4016.64 j/g.°C
The amount of substance present in a certain object with a given half-life in terms of h can be expressed through the equation,
A(t) = (A(o))(0.5)^(t/h)
where A(t) is the amount of substance after t years and A(o) is the original amount. In this item we are given that A(t)/A(o) is equal to 0.89. Substituting the known values,
0.89 = (0.5)(t / 5730 years)
The value of t from the equation is 963.34 years.
<em>Answer: 963 years</em>
Answer:
See explanation
Explanation:
A flammable solvent refers to a solvent that catches fire easily. The precautions to be taken when working with flammable solvents are;
1) heat the solvent at a low to medium hot plate setting.
2) if you need to boil the solvent, use a condenser rather than a flask or beaker without a cover.
3) make sure that the hotplate is larger than the vessel containing the mixture that is being heated.
4) do not use strong oxidizing agents
Answer:
We have to take 37.5 mL of a 0.400 M solution
Explanation:
Step 1: Data given
Stock volume = 100 mL = 0.100L
Stock concentration 0.400 M
Volume of solution he wants to make = 100 mL = 0.100L
Concentration of solution he wants to make = 0.150 M
Step 2: Calculate the volume of 0.400 M CuSO4 needed
C1*V1 = C2*V2
⇒with C1 = the stock concentration = 0.400M
⇒with V1 = the volume of the stock = TO BE DETERMINED
⇒with C2 = the concentration of the solution he wants to make = 0.150 M
⇒with V2 = the volume of the solution made = 0.100 L
0.400 M * V1 = 0.150M * 0.100L
V1 = (0.150M*0.100L) / 0.400 M
V1 = 0.0375 L = 37.5 mL
We have to take 37.5 mL of a 0.400 M solution
Generally speaking, organic molecules tend to dissolve in solvents that have similar physical properties. A good rule of thumb is that "like dissolves like". Meaning, polar compounds can dissolve polar compounds and nonpolar compounds can dissolve nonpolar compounds.
To apply this to the current problem, we are told that the brushes are being cleaned with vegetable oil or mineral oil. In this case, the oils are used as solvents. In order for these solvents to be effective, the compounds they are trying to dissolve must be similar in structure and properties to other oils. Therefore, vegetable oil or mineral oil will be most effective in removing oil-based paints, as these will have the similar properties needed to dissolve in the oil solvents.
