Answer:
The new volume of the gas when the pressure is changed to 2.84 atm is 822.82 mL.
Explanation:
Boyle's law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." That is, if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
or
P * V = k
Assuming that you have a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment, by varying the volume of gas to a new value V2, then the pressure will change to P2, and it will be fulfilled:
P1*V1= P2*V2
In this case:
- P1= 9.86 atm
- V1= 237 mL
- P2= 2.84 atm
- V2=?
Replacing:
9.86 atm* 237 mL= 2.84 atm*V2
and solving you get:
V2= 822.82 mL
<u><em>The new volume of the gas when the pressure is changed to 2.84 atm is 822.82 mL.</em></u>
Answer:
The answer to your question is: 63.5 u
Explanation:
Data Percent Molecular mass
63 Cu 69.2% 62.9296 u
65Cu 30.8% 64.9278 u
Average atomic mass = ?
Formula
Average atomic mass = (0.692 x 62.9296) + (0.308 x 64.9278)
= 43.55 + 19.99
= 63.5 u
Answer:
Molality = 0.0862 mole/kg
Explanation:
Molality = (number of moles of solute)/(mass of solvent in kg)
Number of moles of solute = (mass of Creatinine in the blood sample)/(Molar mass of Creatinine)
To obtain the mass of creatinine in 10 mL of blood. We're told that 1 mg of Creatinine is contained in 1 decilitre of blood.
1 decilitre = 100 mL
1 mg of Creatinine is contained in 100 mL of blood
x mg of Creatinine is contained in 10 mL of blood.
x = (1×10/100) = 0.1 mg = 0.0001 g
Molar mass of Creatinine (C₄H₇N₃O) = 113.12 g/mol
Number of moles of Creatinine in the 10 mL blood sample = (0.0001/113.12) = 0.000000884 moles
Mass of 10 mL of blood = density × volume = 1.025 × 10 = 10.25 mg = 0.01025 g = 0.00001025 kg
Molality of normal creatinine level in a 10.0-ml blood sample = (0.000000884/0.00001025)
Molality = 0.0862 moles of Creatinine per kg of blood.
Hope this Helps!!!
Answer:
Lower the pH slightly
Explanation:
A buffer is defined as the mixture between weak acid and conjugate base. In the problem, acetic acid is the weak acid and sodium acetate the conjugate base.
When a strong acid as perchloric acid is added to a buffer, <em>the pH decreases slightly</em> because the acid reacts with conjugate base producing weak acid and not affecting directly the concentration of H⁺ ions.
