Answer:
2.29 × 10⁻⁴ moles
Explanation:
Data Given:
Chemical Symbol of Zinc = Zn
Mass of Zinc = 1.5 × 10⁻² g
A.Mass of Zinc = 65.38 g.mol⁻¹
The Atomic Mass of Zinc can be obtained from periodic table. While, the mass is given in statement. Also, it is important to know the symbol because some elements exist in diatomic or polyatomic forms i.e. H₂, N₂, S₈ e.t.c.
Mole is the unit used to calculate the amount of substance. It has following general forms,
Mole = Mass / M.Mass
And,
Mole = # of Particles / 6.022 × 10²³ particles.mol⁻¹
Hence, using first equation,
Moles = 1.5 × 10⁻² g / 65.38 g.mol⁻¹
Moles = 2.29 × 10⁻⁴ moles
Answer:
The solution will not form a precipitate.
Explanation:
The Ksp of PbI₂ is:
PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)
Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] <em>Concentrations in equilibrium</em>
When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:
[I⁻] = 0.00345M × (328mL / (328mL+703mL) =<em> 1.098x10⁻³M</em>
[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) =<em> 5.469x10⁻³M</em>
<em />
Q = [I⁻]²[Pb²⁺] <em>Concentrations not necessary in equilibrium</em>
If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.
Replacing:
Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹
As Q < Ksp, the solution is not saturated and <em>will not form a precipitate</em>.
Answer:
The C14 would be found in Glyceraldehyde-3-Phosphate, and the O18 would also be found in the same molecule
Explanation:
Isotopic labeling is a common method for deducing reaction mechanism in chemistry.
In photosynthesis, the oxygen in the found in Glyceraldehyde-3-Phosphate comes from the carbon dioxide, This is also finally found in the glucose produced at the end of the cycle.
The oxygen in water is released into the atmosphere as the oxygen molecule.
Therefore, the C14 would be found in Glyceraldehyde-3-Phosphate, and the O18 would also be found in the same molecule.
Answer:
i believe it's c but I'm not one hundred percent sure
Explanation:
<em>Answer:</em>
![\boxed{N_t =0.1953125 \: grams}](https://tex.z-dn.net/?f=%5Cboxed%7BN_t%20%3D0.1953125%20%5C%3A%20grams%7D)
Explanation:
<em>Formula:</em>
![N_t = N_0( { \frac{1}{2} })^{ \frac{t}{t_ \frac{1}{2} } }](https://tex.z-dn.net/?f=N_t%20%3D%20N_0%28%20%7B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%29%5E%7B%20%5Cfrac%7Bt%7D%7Bt_%20%5Cfrac%7B1%7D%7B2%7D%20%7D%20%7D%20)
Where Nt is the amount of substance remaining after time t,
N0 is the amount of substance at time t = 0,
t is the time at which we have to find out how much substance disintegrated from t = 0 to t = t
& t_1/2 is the half life corresponding radioactive sample.
<em>Given:</em>
N0 = 200g
t= 40 days
t1/2 = 4 days
<em>To find:</em>
Nt =?
<em>Solution:</em>
Substituting given data in above formula,
![N_t = N_0( { \frac{1}{2} })^{ \frac{t}{t_ \frac{1}{2} } } \\ N_t = 200( { \frac{1}{2} })^{ \frac{40}{4 } } \\ N_t = 200 \times 0.0009765625 \\ \boxed{N_t =0.1953125 \: grams}](https://tex.z-dn.net/?f=N_t%20%3D%20N_0%28%20%7B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%29%5E%7B%20%5Cfrac%7Bt%7D%7Bt_%20%5Cfrac%7B1%7D%7B2%7D%20%7D%20%7D%20%5C%5C%20N_t%20%3D%20200%28%20%7B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%29%5E%7B%20%5Cfrac%7B40%7D%7B4%20%7D%20%7D%20%5C%5C%20N_t%20%3D%20200%20%5Ctimes%200.0009765625%20%5C%5C%20%5Cboxed%7BN_t%20%3D0.1953125%20%20%5C%3A%20grams%7D%20)
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