Explanation:
Start with a balanced equation.
2H2 + O2 → 2H2O
Assuming that H2 is in excess, multiply the given moles H2O by the mole ratio between O2 and H2O in the balanced equation so that moles H2O cancel.
5 mol H2O × (1 mol O2/2 mol H2O) = 2.5 mol O2
Answer: 2.5 mol O2 are needed to make 5 mol H2O, assuming H2 is in excess.
Answer:
Li and H
Explanation:
2Li(s)+2H2O(i)→2LiOH(aq)+H2(g) is full balanced
Ionization energy is the energy required to remove the
outermost electron from one mole of gaseous atom to produce 1 mole of gaseous
in to produce a charge of 1. The greater the ionization energy, the greater is
the chance f the electron to be removed from the nucleus. In this casse, Radium
has the largest ionization energy.
Answer: from what i can tell it looks like a heterogeneous mixture
Explanation:
heterogeneous is multiple different components, homogenous includes only the same component
Answer:
(a) 
(b) 
Explanation:
Hello,
In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

Thus, since the final pressure is 3.60 bar, we can write:

The moles of helium could be computed via solver as:

Or algebraically:

In such a way, the volume of the compartment B is:

Finally, he mole fraction of He is:

Regards.