It's difficult to write it down, but I'll attach you a good example of hydroboration of indene. I hope you'll find it helpful.
Answer:
B. double-replacement RXN
Explanation:
more specifically, this is a precipitation rxn.
Answer:
23.92 g
Explanation:
Molar mass of H2SO4 = (2×1)+32+(16×4)= 2+32+48= 82g/mol
H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
I mole of H2SO4 = 2 moles of NaOH
24.5/82 = 24.5/82 × 2
= 0.598 moles of NaOH will neutralize
Mass= mole× molar mass
Molar mass of NaOH= 23+16+1 = 40g/mol
Mass= 0.598 × 40 = 23.92g of NaOH
Answer:
hope you like and pls follow me
Explanation:
digestive and circulatory system are most important in breaking down food and distributing nutrients throughout the body.
Answer:
4L
Explanation:
To obtain the volume of O2 at stp, first, we need to determine the number of mole of O2.
From the question given above,
Mass of O2 = 5.72g
Molar Mass of O2 = 32g/mol
Number of mole =Mass/Molar Mass
Number of mole of O2 = 5.72/32
Number of mole of O2 = 0.179 mole
Now, we can calculate the volume of O2 at stp as follow:
1 mole of a gas occupy 22.4L at stp.
Therefore, 0.179 mole of O2 will occupy = 0.179 x 22.4 = 4L
Therefore, the volume occupied by the sample of O2 is 4L