The piece of jewelry contains 5.08x10 21to the power dozen Ag atoms.
Answer: Mass of
produced in this reaction was 6.56 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
![CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)](https://tex.z-dn.net/?f=CaCO_3%28s%29%2B2HCl%28aq%29%5Crightarrow%20H_2O%28l%29%2BCO_2%28g%29%2BCaCl_2%28aq%29)
Mass or reactants = Mass of
+ mass of
= 16.00 + 64.80 = 80.80 g
Mass of products = mass of aqueous solution + mass of
+ = 74.24 + x g
Mass or reactants = Mass of products
80.80 g = 74.24 + x g
x = 6.56 g
Thus mass of
produced in this reaction was 6.56 grams
For the following question(s), consider a 4% starch solution and a 10% starch solution separated by a semipermeable membrane.
Which of the following also occurs in this system?
There is a net flow of water from the 4% starch solution into the 10% starch solution
Answer:
5.0 x 10⁹ years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of K-40 = 1.251 × 10⁹ years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of (K-40) ([A₀] = 100%).
[A] is the remaining concentration of (K-40) ([A] = 6.25%).
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.
∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.
Answer:
Oxidation number of F2O = 0−(−1×2)
State of oxygen will be=+2