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>
Answer:
1.22 mL
Explanation:
Let's consider the following balanced reaction.
2 AgNO₃ + BaCl₂ ⇄ Ba(NO₃)₂ + 2 AgCl
The molar mass of silver chloride is 143.32 g/mol. The moles corresponding to 0.525 g are:
0.525 g × (1 mol/143.32 g) = 3.66 × 10⁻³ mol
The molar ratio of AgCl to BaCl₂ is 2:1. The moles of BaCl₂ are 1/2 × 3.66 × 10⁻³ mol = 1.83 × 10⁻³ mol.
The volume of 1.50 M barium chloride containing 1.83 × 10⁻³ moles is:
1.83 × 10⁻³ mol × (1 L/1.50 mol) = 1.22 × 10⁻³ L = 1.22 mL
When we can get the Kinetic energy from this formula KE= 1/2 M V^2and we can get the potential energy from this formula PE = M g H
we can set that the kinetic energy at the bottom of the fall equals the potential energy at the top so, KE = PE
1/2 MV^2 = M g H
1/2 V^2 = g H
when V is the velocity, g is an acceleration of gravitational force (9.8 m^2/s) and H is the height of the fall (8 m).
∴ v^2 = 2 * 9.8 * 8 = 156.8
∴ v= √156.8 = 12.5 m/s
Answer:
44Kj
Explanation:
These are the equations for the reaction described in the question,
Vaporization which can be defined as transition of substance from liquid phase to vapor
H2(g)+ 1/2 O2(g) ------>H2O(g). Δ H
-241.8kj -------eqn(1)
H2(g)+ 1/2 O2(g) ------>H2O(l).
Δ H =285.8kj ---------eqn(2)
But from the second equation we can see that it moves from gas to liquid, we we rewrite the equation for vaporization of water as
H2O(l) ------>>H2O(g)---------------eqn(3)
But the equation from eqn(2) the eqn does go with vaporization so we can re- write as
H2O ------> H2(g)+ 1/2 O2(g)
Δ H= 285.8kj ---------------eqn(4)
To find Delta h of the vaporization of water at these conditions, we sum up eqn(1) and eqn(4)
Δ H=285.8kj +(-241.8kj)= 44kj
Answer:
H₃PO₄/H₂PO₄⁻ and HCO₃⁻/CO₃²⁻
Explanation:
An acid is a proton donor; a base is a proton acceptor.
Thus, H₃PO₄ is the acid, because it donates a proton to the carbonate ion.
CO₃²⁻ is the base, because it accepts a proton from the phosphoric acid.
The conjugate base is what's left after the acid has given up its proton.
The conjugate acid is what's formed when the base has accepted a proton.
H₃PO₄/H₂PO₄⁻ make one conjugate acid/base pair, and HCO₃⁻/CO₃²⁻ are the other conjugate acid/base pair.
H₃PO₄ + CO₃²⁻ ⇌ H₂PO₄⁻ + HCO₃⁻
acid base conj. conj.
base acid
