Answer:
see explanation below
Explanation:
To do this exercise, we need to use the following expression:
P = nRT/V
This is the equation for an ideal gas. so, we have the temperature of 22 °C, R is the gas constant which is 0.082 L atm / mol K, V is the volume in this case, 5 L, and n is the moles, which we do not have, but we can calculate it.
For the case of the oxygen (AW = 16 g/mol):
n = 30.6 / 32 = 0.956 moles
For the case of helium (AW = 4 g/mol)_
n = 15.2 / 4 = 3.8 moles
Now that we have the moles, let's calculate the pressures:
P1 = 0.956 * 0.082 * 295 / 5
P1 = 4.63 atm
P2 = 3.8 * 0.082 * 295 / 5
P2 = 18.38 atm
Finally the total pressure:
Pt = 4.63 + 18.38
Pt = 23.01 atm
In this question, you are given the NaOH volume but asked for concentration.
Don't forget that for every 1 mol of NaOH there will be 1 mol OH- ion, but for every 1 mol of H2SO4 there will be 2 mol of H- ion.
To neutralize you need the same amount of OH- and H+, so the equation should be:
OH-= H+
<span>35.50cm3 * x*1= 25cm3* 0.2mol/dm3 *2
</span>x= 10/35.5 mol/dm3= 0.2816/dm3
<span>Answer:
Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass.
râšM = constant
Therefore for two gases the ratio rates is given by:
r1 / r2 = âš(M2 / M1)
For Cl2 and F2:
r(Cl2) / r(F2) = âš{(37.9968)/(70.906)}
= 0.732 (to 3.s.f.)</span>
Answer:
a) find attached image 1
b) find attached image 2
Explanation :
The more stable radical is formed by a reaction with smaller bond dissociation energy.
since the bond dissociation for cleavage of the bond to form primary free radical is higher, more energy must be added to form it. This makes primary free radical higher in energy and therefore less stable than secondary free radical.
Answer:
lignands, the central atom/metal ion
Explanation:
