near what like open flames? if so false unless it clarifies the solution requires alcohol then its a no go
Since orbital period depends on how far you are from the sun, planets closer to the sun have a orbital period less than one earth year.
These planets are Mercury and Venus
Answer:
6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.
Explanation:
![Rate = k[AB]^2](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BAB%5D%5E2)
The order of the reaction is 2.
Integrated rate law for second order kinetic is:
Where,
is the initial concentration = 1.50 mol/L
is the final concentration = 1/3 of initial concentration =
= 0.5 mol/L
Rate constant, k = 0.2 L/mol*s
Applying in the above equation as:-


<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>
Some of the reactants or the products are in the gaseous phase.
Answer:
1.327 g Ag₂CrO₄
Explanation:
The reaction that takes place is:
- 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)
First we need to <em>identify the limiting reactant</em>:
We have:
- 0.20 M * 50.0 mL = 10 mmol of AgNO₃
- 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄
If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.
Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:
- 4 mmol K₂CrO₄ *
= 1326.92 mg Ag₂CrO₄
- 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄