Answer is: <span>the volume of water after the solid is added</span> is 4.5 ml.
d(gold) = 8.0 g/cm³; density of gold.
m(gold) = 4 g; mass of gold.
V(gold) = m(gold) ÷ d(gold); volume of gold.
V(gold) = 4 g ÷ 8 g/cm³.
V(gold) = 0.5 cm³ = 0.5 ml.
V(water) = 4.00 ml = 4.00 cm³.
V(flask) = V(gold) + V(water).
V(flask) = 0.5 cm³ + 4 cm³.V = 4.5 cm³.
To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:
P1V1 = P2V2,
Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:
(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr
The final pressure exerted by the gas would be 0.593 torr.
Hope this helps!
Answer:
0.48 V
Explanation:
Zn(s) ------------> Zn^2+(aq) + 2e. Oxidation half equation (-0.76V)
Co^2+(aq) + 2e-----------> Co(s). Reduction half equation (-0.28)
Zn(s) + Co^2+(aq) -------------> Zn^2+(aq) + Co(s) overall redox equation
Zinc is the anode while cobalt is the cathode.
E°cell= E°cathode - E°anode
E°cell= -0.28-(-0.76)= 0.48 V
Answer:
-255.4 kJ
Explanation:
The free energy of a reversible reaction can be calculated by:
ΔG = (ΔG° + RTlnQ)*n
Where R is the gas constant (8.314x10⁻³ kJ/mol.K), T is the temperature in K, n is the number of moles of the products (n =1), and Q is the reaction quotient, which is calculated based on the multiplication of partial pressures by the partial pressure of the products elevated by their coefficient divide by the multiplication of the partial pressure of the reactants elevated by their coefficients.
C₂H₂(g) + 2H₂(g) ⇄ C₂H₆(g)
Q = pC₂H₆/[pC₂H₂ * (pH₂)²]
Q = 0.261/[8.58*(3.06)²]
Q = 3.2487x10⁻³
ΔG = -241.2 + 8.314x10⁻³x298*ln(3.2487x10⁻³)
ΔG = -255.4 kJ
Answer : The retention time is, 20 min
Explanation :
Retention time : It is defined as the amount of time a compound spends on the column after it has been injected.
Formula of retention time is:
Given:
Distance from injection point to center of peaks = 10 cm
Chart recorded speed = 0.5 cm/min
Now put all the given values in the above formula, we get:
Retention time = 20 min
Thus, the retention time is, 20 min
