Kr look on periodic table it's krypton elements
Answer:
I believe that it is the 2nd option.
Explanation:
My reasonings are because C4H10O has 7 isomers. In which 4 are alcohol and the other 3 are ether.
The first option is ethers, specifically ethoxyethane.
The third option is ethers, specifically 1-methoxypropane.
The fourth option is an alcohol, specifically 1- butanol.
Therefore, leads us to the 2nd option that it is NOT an isomer of C4H10O
francium , in the Periodic table the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. making helium is the smallest element, and francium the largest.
Answer:
The new volume is 5.37 L
Explanation:
Step 1: Data given
Initial volume = 10.0 L
Initial temperature = 22.0 °C
Initial pressure = 1.00 atm
Final temperature = 202 °C
Final pressure = 3.00 atm
Step 2: Calculate final volume
(P1*V1)/T1 = (P2*V2)/T2
⇒ with P1 = The initial pressure = 1.00 atm
⇒ with V1 = The initial volume = 10.0 L
⇒ with T1 = The initial temperature = 22 °C = 295 Kelvin
⇒ with P2 = The final pressure = 3.00 atm
⇒ with V2 = The final volume = TO BE DETERMINED
⇒ with T2 = The final temperature = 202 °C = 475 Kelvin
(1.00 * 10.0) / 295 = (3.00 * V2) / 475
10 / 295 = 3V2/ 475
3V2 = 4750/295
V2 = 5.37 L
The new volume is 5.37 L
Answer:
Please see the complete formt of the question below
Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.
MnO₂(s) + HCl(aq) → MnCl₂(aq) + H₂O(l) + Cl₂(g)
According to the above reaction, determine the limiting reactant when 5.6 moles of MnO₂ are reacted with 7.5 moles of HCl.
The answer to the above question is
The limiting reactant is the MnO₂
Explanation:
To solve this, we note that one mole of MnO₂ reacts with one mole of HCl to produce one mole of MnCl₂, one mole of H₂O and one mole of Cl₂
Molar mass of MnO₂ = 86.9368 g/mol
Molar mass of HCl = 36.46 g/mol
From the stoichiometry of the reaction, 5.6 moles of MnO₂ will react with 5.6 moles of HCl to produce 5.6 moles of H₂O and 5.6 moles of Cl₂
However there are 7.5 moles of HCL therefore there will be an extra 7.5-5.6 or 1.9 moles of HCl remaining when the reaction is completed