Answer:
2450 cm3
Explanation:
Volume of cylinder = V=πr2h
2.45L = 2450mL
1mL = 1 cm cubed
2450mL = 2450 cm cubed
Explanation:
a) Using the provided information about the density of gold, the sample size, thickness, and the following equations and comersion factors, find the area of the gold leaf:

Gold 

First, find the volume of the sample and then find the area of the sample.


b. Using the provided information from part
), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :


Answer:
Increasing the volume of the vessel
Explanation:
By the Le Chatelier's principle, if a system in equilibrium suffer a variation that disturbs the equilibriu, the reaction shift in the way to minimize the pertubation and re-establish the equilibrium.
For a variation in pressure, when it increases, the reaction shift for the smallest of gas volume, and if decreases, the reaction will shift for the large gas volume. So, for the reaction given, the products have the large amount of gas, so by decreasing the pressure, more products will be formed, and the amount of NH₄HS will reduce. To decrease the pressure, we can increase the volume of the vessel: for the ideal gas equation (PV= nRT), pressure and volume are indirectly proportional.
Answer:
V2 = 6616 L
Explanation:
From the question;
Initial volume = 40L
Initial Pressure, P1 = 159atm
Initial Temperature T1 = 25 + 273 = 298K (Upon converting to Kelvin unit)
Final Volume, V2 = ?
Final Pressure, P2 = 1 atm
Final Temperature T2 = 37 + 273= 310K (Upon converting to Kelvin unit)
These quantities are related by the equation;
P1V1 / T1 = P2V2 / T2
V2 = T2 * P1 * V1 / T1 * P2
V2 = 310 * 159 * 40 / (298 * 1)
V2 = 6616 L
Answer:
296.1 day.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).
t is the time of the reaction (t = ??? day).
a is the initial concentration of Ir-192 (a = 560.0 dpm).
(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).
<em>∴ kt = lna/(a-x)</em>
(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).
(9.365 x 10⁻³ day⁻¹)(t) = 2.773.
<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>