Answer:
5. the scattering of α particles by a metal foil
Explanation:
This is the classical Rutherford's experiment in which he bombarded a thin foil of gold with alpha particles which are positively charged helium nucleus.
He did observed that most of the particles passed through the foil relatively undeflected or if they were deflected it was by a very small angle.
Once in a while the alpha particle rebounded completely. An analogy is the one typically mentioned that it was as if we throw a ball at a piece of paper and it rebounds toward us.
This observations led Rutherford to conclude that the nucleus of the atom is very small positely charged and that the atom is relatively empty with electrons of very small masses. His model is referred as the Plum Pudding model and later Bohr modified it to the planetary model.
The chemical reaction for this is:
2 C2H6 + 7 O2 => 4 CO2 + 6 H2O
Solving for CO2 with each reactant will give:
21.0 g C2H6 x (1 mol C2H6/30.08 g C2H6) x (6 mol H2O/2
mol C2H6) x (18 g H2O/1 mol H2O) = 37.70 g H2O
110 g O2 x (1 mol O2/32.00 g O2) x (6 mol CO2/7 mol O2) x
(18 g H2O/1 mol H2O) = 53.04 g H2O
Since the amount of H2O in C2H6 is lower therefore C2H6
is the limiting reactant and the maximum amount of water is only 38 g H2O (2 significant digits)
ANswer:
38 g water
The stock solution has a volume of 0.208L. When diluted, a stock solution with a concentration of 0.30 m clo2 becomes a 0.20 m stock solution with a volume of 0.050.
A three-dimensional space's occupied volume is measured. It is frequently measured numerically with SI-derived units (such the cubic metre and litre) or with other imperial units (such as the gallon, quart, cubic inch). Volume and length are related in how they are defined (cubed).
A highly concentrated stock solution, often prepared as a 10x concentrated solution, can be described as a stock solution since it can be diluted to create a working solution. In order to prepare complex solutions with numerous constituents, stock solutions can also be utilized as a component.
0.3M*v = 0.125M*0.50L
V=0.208L
Learn more about stock solution here
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It is the first option bc if it absorbed heat it it will pressurize and could explode.
Answer:
2621.75 j heat is required to increase the temperature 25.5°C to 46°C.
Explanation:
Given data:
Mass of sample = 142.1 g
Initial temperature = 25.5°C
Final temperature = 46°C
Specific heat capacity of Al = 0.90 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 46°C - 25.5°C
ΔT = 20.5°C
Q = 142.1 × 0.90 J/g.°C × 20.5°C
Q = 2621.75 j
Thus, 2621.75 j heat is required to increase the temperature 25.5°C to 46°C.
