Here we assume that disintegration has occur so the normal t½ formular will not be used
mathematically we..⅛×5600 =700
therefore it is about 700years option A
The tool to measure the liquid is a measuring cylinder.
The formula for molality---> m = moles solute/ Kg of solvent
the solute here is NH₃ because it's the one with less amount. which makes water the solvent.
1) let's convert the grams of NH₃ to moles using the molar mass
molar mass of NH₃= 14.0 + (3 x 1.01)= 17.03 g/ mol
15.0 g (1 mol/ 17.03 g)= 0.881 mol NH₃
2) let's convert the grams of water into kilograms (just divide by 1000)
250.0 g= 0.2500 kg
3) let's plug in the values into the molality formula
molality= mol/ Kg---> 0.881 mol/ 0.2500 kg= 3.52 m
> 2,000
mL of a 5.0 × 10–5% (w/v) sucrose solution
5.0 × 10–3
g/mL * 2000 mL * (1 mol / 342.30 g) = 0.0292 mol
<span>
> 2,000 mL of a 5.0 ppm sucrose solution</span>
5 grams /
1000000 mL * 2000 mL* (1 mol / 342.30 g) = 0.0000292 mol
<span>
> 20 mL of a 5.0 M sucrose solution </span>
5.0 M *
0.020 L = 0.1 mol
Answer:
<span>2,000 mL
of a 5.0 ppm sucrose solution</span>
Answer:
1.51 X 10^23 ions
Explanation:
The number of ions in 17.1 gm of aluminum sulphate Al2 (SO4)3 =….. [Molar mass of Al2 (SO4)3 = 342 gm]
in one molecule of Al2(SO4)3 there are 5 ions 2 aluminum and 3 sulfate ions
in 2 molecules there are 2X5= 10 ions
in 10 molecules there are 10X5 = 50 ions
molar mass of Al2(SO4)3 = (2 X 26.98) +( 3 X 32.1) + (3 X 4 X 16.0 ) =342.gms = 17.1/342 =0.0500 moles
1 mole =6.02 X 10^23 molecules ( see Avogadros number)
0.0500 moles = 0.0500 X 6.02 X 10^23 molecules =
0.301 X 10^23 molecules = 3.01 X 10^22 molecules
We determined that each molecule of Al2(SO4)3 has 5 ions
so 3.01 X10^22 molecules have 5 X 3.01 X 10^22 ions =
15.05 X 10^22 ions = 1.51 X 10^23 ions
