Answer: gas particles that are in constant motion and exhibit perfectly elastic collisions
Explanation:
Answer is: the molar mass of gas is 82.83 g/mol.
m(gas) = 55.50 g; mass of gas.
V(gas) = 15.0 L; volume of gas.
Vm = 22.4 L/mol; molar volume of gas on STP.
n(gas) = V(gas) ÷ Vm.
n(gas) = 15 L ÷ 22.4 L/mol.
n(gas) = 0.67 mol; amount of gas.
M(gas) = m(gas) ÷ n(gas).
M(gas) = 55.5 g ÷ 0.67 mol.
M(gas) = 82.83 g/mol.
Answer:
i) During washing
ii) During Measurement
Explanation:
The two possible steps are :
<u>i) During washing </u>: during the washing method the residue may be not completely dried out and this residue ( water) will add up to the final product ( copper yield ) and this kind of error is called human error.
<u>ii) During measurement </u>: If the weighing instrument is faulty there might be addition in value of the final copper yield which will see the final yield value > 100% . this error occurs when the initial value and final value is been weighed
Answer:

Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
Given: Kp =
Temperature = 25°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
T = (25 + 273.15) K = 298.15 K
R = 0.082057 L atm.mol⁻¹K⁻¹
Δn = (2)-(2+1) = -1
Thus, Kc is:


<u>Answer:</u> The fraction of the rock that is still composed of potassium-40 is 0.25
<u>Explanation:</u>
To calculate the fraction of the rock that is still composed of K-40, we use the formula:

where,
a = amount of reactant left after n-half lives
= Initial amount of the reactant
n = number of half lives = 2
Putting values in above equation, we get:

Hence, the fraction of the rock that is still composed of potassium-40 is 0.25