Answer:
Final Volume = 5.18 Liters
Explanation:
Initial Condition:
P1 = 789 mm Hg x (1/760) atm /mm Hg = 1.038 atm
T1 = 22° C = 273 + 22 = 295 K
V1 = 4.7 L
Final Condition:
P2 = 755 mm Hg x (1/760) atm /mm Hg = 0.99 atm
T2 = 37° C = 273 + 37 = 310 K
V2 = ?
Since, (P1 x V1) / T1 = (P2 x V2) / T2,
Therefore,
⇒ (1.038)(4.7) / 295 = (0.99)(V2) / 310
⇒ V2 = 5.18 L (Final Volume)
Its structure is Au with one dot on top
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)
What is the percent by mass of sodium in Na2SO4? total mass of element in compound molar mass of compound Use %Element x 100
