No because it changes its composition to ozone
Answer:
Hello There!!
Explanation:
The atom is still 24Na.
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Answer:
Rubidium-85=61.2
Rubidium-87=24.36
Atomic Mass=85.56 amu
Explanation:
To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.
<u>Rubidium-85 </u>
This isotope has an abundance of 72%.
Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.
- 72/100= 0.72 or 72.0 --> 7.2 ---> 0.72
Multiply the mass of the isotope, which is 85, by the abundance as a decimal.
- mass * decimal abundance= 85* 0.72= 61.2
Rubidium-85=61.2
<u>Rubidium-87</u>
This isotope has an abundance of 28%.
Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.
- 28/100= 0.28 or 28.0 --> 2.8 ---> 0.28
Multiply the mass of the isotope, which is 87, by the abundance as a decimal.
- mass * decimal abundance= 87* 0.28= 24.36
Rubidium-87=24.36
<u>Atomic Mass of Rubidium:</u>
Add the two numbers together.
- Rb-85 (61.2) and Rb-87 (24.36)
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
