Answer: A) Storing experimental samples
Explanation:
It is a common piece of laboratory glassware that can be made of glass or plastic and is opened at the top and closed at the bottom.
It cannot be used for measurements because there is no graduation indicating the volume.
Althought it can contain extra chemicals left over from an experiment, it is not the main proposal of the glassware that is to store samples.
It cannot be used in a microscope and the object for that is a microscope slide.
With the lack of these 3 factors you throw off your health triangle and your health is unbalanced. When you eat well you can get vitamins such as "E" and "C". These vitamins help build up immune cells that help build the immune system. When a person does not get enough sleep that person is more open to viruses and bacteria. Getting the recommended vaccines help you to fight off possible diseases or illnesses. They make you immune system stronger so you can be healthier.
Answer:
Q sln = 75.165 J
Explanation:
a constant pressure calorimeter:
∴ m sln = m Ba(OH)2 + m HCl
∴ molar mass Ba(OH)2 = 171.34 g/mol
∴ mol Ba(OH)2 = (0.06 L)(0.3 mol/L) = 0.018 mol
⇒ mass Ba(OH)2 = (0.018 mol)(171.34 g/mol) = 3.084 g
∴ molar mass HCl = 36.46 g/mol
∴ mol HCl = (0.06 L)(0.60 mol/L) = 0.036 mol
⇒ mass HCl = (0.036 mol)(36.46 g/mol) = 1.313 g
⇒ m sln = 3.084 g + 1.313 g = 4.3966 g
specific heat (C):
∴ C sln = C H2O = 4.18 J/g°C
∴ ΔT = 26.83°C - 22.74°C = 4.09°C
heat absorbed (Q):
⇒ Q sln = (4.3966 g)(4.18 J/g°C)(4.09°C)
⇒ Q sln = 75.165 J
I am sure, the answer is variant B.
Answer:
a=28600J; b=90.6 J/K; c=402 torr
Explanation:
(a) considering the data given
Vapour pressure P1 =0 at Temperature T1 = 42.43˚C,
Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)
Using the Clausius-Clapeyron Equation
ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
In 760/140 = ΔH/8.314 J/mol/K × (1/315.58K -- 1/273.15K)
ΔH vap= +28.6 kJ/mol or 28600J
(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.
Since ΔG at boiling point is zero,
ΔS =(ΔH°vap/Τb)
ΔS = 28600 J/315.58 K
= 90.6 J/K
(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
ln P298 K/1 atm = 28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)
P298 K = 0.529 atm
= 402 torr
