I think the answer is ‘repulsion’
Answer:
son 12.6 gramos de HF
Explanation:
Tienes que saber qual es el reactor limitante en este caso es fluoruro con los 20 gramos puedes producer .631 mol qual son 12.6 gramos
Answer:
molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x 1 mole CH4 = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.
Explanation:
not to certain if this is right or not.. but hope it helps!
Answer:
just use the tongs and put it on a plate
Explanation:
3Na2O(at) + 2Al(NO3)3(aq) —> 6NaNO3(aq) + Al2O3(s)
This is a double replacement reaction and NaNO3 is aqueous because Na is an alkali metal, plus nitrate is in the solution. Both of these are soluble. Al2O3 is not soluble because it does not contain any element that is soluble and is hence the precipitate.
Hope this helped!
