Answer:
The sugar particles are so small that they have dissolved in the water, and can easily pass through the filter.
Answer: b.) they tend to lose electrons to gain stability
Explanation:
16.4 grams is the mass of solute in a 500 mL solution of 0.200 M
.
sodium phosphate
Explanation:
Given data about sodium phosphate
atomic mass of Na3PO4 = 164 grams/mole
volume of the solution = 500 ml or 0.5 litres
molarity of sodium phosphate solution = 0.200 M
The formula for molarity will be used here to know the mass dissolved in the given volume of the solution:
The formula is
molarity = 
putting the values in the equation, we get
molarity x volume = number of moles
0.200 X 0.5= number of moles
number of moles = 0.1 moles
Atomic mass x number of moles = mass
putting the values in the above equation
164 x 0.1 = 16.4 grams
16.4 grams of sodium phosphate is present in 0.5 L of the solution to make a 0.2 M solution.
Answer:
1 .
2.
Explanation:
The more stable the ionic compound, the more is it lattice energy.
- The more the charge on the cation and the anion, the greater is the lattice energy.
- The less the size of the cation and the anion, the greater is the lattice energy.
Scandium oxide (
) is an oxide in which
behaves as cation and
behaves as anion.
The compounds which has higher lattice energy than scandium oxide are:
1 .
This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation
is smaller than
. Thus, this corresponds to higher lattice energy.
2.
This is because the charge on the cation
is greater than that of
and also the size of the cation
is smaller than
. Thus, this corresponds to higher lattice energy.
Answer:
P2=0.385atm
Explanation:
step one:
Given that the temperature T1= 60 Celcius
we can convert this to kelvin by adding 273k to 60 Celcius
we have T1= 333k
pressure P1= 0.470 atm
step two:
we know that the standard temperature is T2= 273K
Applying the temperature and pressure relationship we have
P1/T1=P2/T2
substituting our given data we have
0.47/333=P2/273
cross multiply we have
P2= (0.47*273)/333
P2= 128.31/333
P2=0.385 atm