The balanced equation for the above reaction is as follows;
CaCO₃ + 2HCl ----> CaCl₂ + H₂O + CO₂
stoichiometry of CaCO₃ to HCl is 1:2
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP.
volume of 22.4 L occupied by 1 mol
therefore 0.56 L occupied by - 0.56 L / 22.4 L/mol = 0.025 mol
number of HCl moles reacted - 0.025 mol
2 mol of HCl reacts with 1 mol of CaCO₃
therefore 0.025 mol reacts with - 0.025/2 = 0.0125 mol
mass of CaCO₃ required - 0.0125 mol x 100 g/mol = 1.25 g
1.25 g of CaCO₃ is required
<h3>Answer:</h3>
Curium-247 <em>i.e.</em> ²⁴⁷₉₆Cm
<h3>Explanation:</h3>
Alpha decay is given by following general equation,
ᵃₓA → ⁴₂He + ᵃ⁻⁴ₓ₋₂B
Where;
A = Parent Isotope
B = Daughter Isotope
ᵃ = Mass Number
ₓ = Atomic Number
Californium-251 is the parent isotope in our case and it has 98 protons (atomic number) and is given as,
²⁵¹₉₈Cf
The alpha decay reaction of Californium-251 will be as,
²⁵¹₉₈Cf → ⁴₂He + ²⁴⁷₉₆B
The symbol for B with atomic number 96 was found to be the atom of Curium (Cm) by inspecting periodic table. Hence, the final equation is as follow,
²⁵¹₉₈Cf → ⁴₂He + ²⁴⁷₉₆Cm
Answer:
14.89 g
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
Answer:
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Explanation:
The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.
To calculate the half-life time we use the following equation:
[At]=[Ai]*e^(-kt)
with [At] = Concentration at time t
with [Ai] = initial concentration
with k = rate constant
with t = time
We want to know the half-life time = the time needed to have 50% of it's initial value
50 = 100 *e^(-8.7 *10^-3 s^- * t)
50/100 = e^(-8.7 *10^-3 s^-1 * t)
ln (0.5) = 8.7 *10^-3 s^-1 *t
t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.