Answer:
V = 3.75 cm³
Explanation:
Density = Mass / Volume
Step 1: Define variables
D = 8.00 g/cm³
M = 30.0 g
V = unknown
Step 2: Substitute and Evaluate for V
8.00 g/cm³ = 30 g/ V
V = 15/4 cm³
V = 3.75 cm³
The results of this experiment gave Rutherford the means to arrive at two conclusions: one<span>, an atom was much more than just empty space and scattered electrons and </span>two<span>, an atom must have a positively charged center that contains most of its mass (which Rutherford termed as the </span>nucleus<span>).
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There is 3800 mmHg in 5 atm
6.02*10^23 is Avagadro's number representing the number of molecules per mole of substance.
Answer:
m(Na3AlF6)=57.6537 kg. Leftover mass will be 52.8644 kg
Explanation:
Al2O3(s)+6NaOH(l)+12HF(g)⟶2Na3AlF6+9H2O(g)
n(Al2O3)=m(Al2O3)/M(Al2O3)=14000/101.96=137.31 mol
n(NaOH)=m(NaOH)/M(NaOH)=59400/40=1485 mol
n(HF)=m(HF)/M(HF)=59400/20.01=2968 mol
To know what is in excess we divide moles by coefficients and compare numbers.
n(Al2O3) = 137.31 mol
n(NaOH)/6=247.5 mol
n(HF)/12=247.33 mol
so NaOH and HF are in excess
So
m(Na3AlF6)=n(Na3AlF6)*M(Na3AlF6)/1000=n(Al2O3)*2*M(Na3AlF6)/1000=57.6537 kg.
Leftover mass will be (n(NaOH)-n(Al2O3)*6)*40 + (n(HF)-n(Al2O3)*12)*20.01=26445.6+26418.8=52864.4 g= 52.8644 kg