Answer:
No photoelectric effect is observed for Mercury.
Explanation:
From E= hf
h= Plank's constant
f= frequency of incident light
Threshold Frequency of mercury= 435×10^3/ 6.6×10^-34 × 6.02×10^23
f= 11×10^14 Hz
The highest frequency of visible light is 7.5×10^14. This is clearly less than the threshold frequency of mercury hence no electron is emitted from the mercury surface
Answer:
186.9Kelvin
Explanation:
The ideal gas law equation is PV
=
n
R
T
where
P is the pressure of the gas
V is the volume it occupies
n is the number of moles of gas present in the sample
R is the universal gas constant, equal to 0.0821
atm L
/mol K
T is the absolute temperature of the gas
Ensure units of the volume, pressure, and temperature of the gas correspond to R
( the universal gas constant, equal to 0.0821
atm L
/mol K
)
n
=
3.54moles
P= 1.57
V= 34.6
T=?
PV
=
n
R
T
PV/nR = T
1.57 x 34.6/3.54 x 0.0821
54.322/0.290634= 186.908620464= T
186.9Kelvin ( approximately to 1 decimal place)
First, we will need to find the density of the object, take the mass and divide it by the dispplaced water:
128/424 = 0.302 grams/milliliters
Convert that to kg/m3
We get: 302kg/m3
Divide that to the density of water: 1000kg/m3
302/1000 = 0.302
(thats a pretty darn light weighted metal)
Answer:
24.9%
Explanation:
According to this question, mole fraction of NaCl in an aqueous solution is 0.0927. This means that the mole percent of NaCl in the solution is:
0.0927 × 100 = 9.27%
Let's assume that the solution contains water (solvent) + NaCl (solute), hence, the mole fraction of water will be;
100% - 9.27% = 90.73%
THEREFORE, it can be said that, NaCl contains 0.0927moles while H2O contains 9.073moles
N.B: mole = mass/molar mass
Given the Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol
For NaCl;
0.0927 = mass/58.44
mass = 0.0927 × 58.44
5.42g
For H2O;
9.073 = mass/18.016
mass = 9.073 × 18.016
= 16.35g
Total mass of solution = 16.35g + 5.42g = 21.77g
Mass percent of NaCl = mass of NaCl/total mass × 100
% mass of NaCl = 5.42g/21.77g × 100
= 0.249 × 100
= 24.9%