Answer:
<em>C</em> C2H5OH = 7.598 molal
Explanation:
- molality (m) C2H5OH = n C2H5OH / Kg H2O
∴ V C2H5OH = 50.0 mL
∴ V H2O = 112.7 mL
∴ T = 20 °C
∴ δ C2H5OH (20°C) = 0.789 g/mL
∴ δ H2O (20°C) = 1.00 g/mL
∴ Mw C2H5OH = 46.0684 g/mol
⇒ n C2H5OH = (50.0 mL)×(0.789 g/mL)×(mol/46.0684 g) = 0.8563 mol
⇒ mass H2O = (112.7 mL)×(1.00 g/mL)×(Kg / 1000 g) = 0.1127 Kg
⇒ <em>C</em> C2H5OH = 0.8563 mol C2H5OH / 0.1127 Kg H2O
⇒ <em>C</em> C2H5OH = 7.598 molal
Answer:
Follows are the solution to this question:
Explanation:
The creation of four fragments,
(n endpoint) MNQSYGR (c endpoint)
Weight of the molecular = 854.92
Load = 2
= 427.46
(n endpoint)RLVSR (c endpoint)
Weight of the molecular = 629.76
Load = 3
= 209.92
(n endpoint) AATAMASLIK (c endpoint)
Weight of the molecular = 976.20
Load at pH(2). = 2
= 442.50
(n endpoint) IFAWWY (c endpoint)
Weight of the molecular = 885.03
Load= 1
= 885.03
It is apparent that RLVSR is the smallest mass ration for this peptide that is detected in the first place.
Answer:
J.J. Thomson
Explanation:
According to my book and wiki, Historically, the mass of the electron was determined directly from combining two measurements. The mass-to-charge ratio of the electron was first estimated by Arthur Schuster in 1890 by measuring the deflection of "cathode rays" due to a known magnetic field in a cathode ray tube. It was seven years later that J. J.
Please correct me if I'm wrong.
Since the pressure is constant, you can use the equation V₁/T₁=V₂/T₂ and solve for T₂. (T₂=(T₁/V₁)V₂)
since this is a gas law type question, the temperatures need to be in kelvin therefore you need to convert 15°C to kelvin. 15°C+273=288K. The volume are already in L so they don't have to be converted.
now you just need to plug everything in:
T₂=(288K/99L)x1200L
T₂=3491K
The answer is you need if you have a 99L of gas at 288K, you have to increase the temperature to 3491K to get a volume of 1200L if the pressure remains constant.
I hope this helps. Let me know in the comments if anything is unclear.
Answer:
The molar volume of a gas is the volume of one mole of a gas at STP. At STP, one mole (6.02 × 1023 representative particles) of any gas occupies a volume of 22.4 L (Figure below). A mole of any gas occupies 22.4 L at standard temperature and pressure (0°C and 1 atm).
Explanation: