Answer:
The initial volume of Ne gas is 261mL
Explanation:
This question can be answered using Ideal Gas Equation;
However, the following are the given parameters
Initial Pressure = 654mmHg
Finial Pressure = 345mmHg
Final Volume = 495mL
Required
Initial Volume?
The question says that Temperature is constant;
This implies that, we'll make use of Boyle's law ideal gas equation which states;
Where 
represent the initial pressure
represent the final pressure
represent the initial temperature
represent the final temperature
Substitute these values in the formula above;
Divide both sides by 654
(Approximated)
<em>The initial volume of Ne gas is 261mL</em>
Answer:
Explanation:
Ionization energy of hydrogen atom is 13.6 eV . This energy will be provided by energetic proton , the kinetic energy of which is 1000 eV. The kinetic energy of ionized electron is 15.2 eV . Kinetic energy of proton produced from from the ionization of hydrogen or the nucleus of the hydrogen atom is 4.3 eV . All these energy must have come from kinetic energy of initial proton.
So kinetic energy of projectile proton after collision
= 1000 - ( 13.6 + 15.2 + 4.3 ) eV.
= 966.9 eV .
The type of fat that is described above is the trans fat. The trans fat is artificially made and these fat contains partially hydrogenated oils which is similar to the statement above. The reason why they are added to oil like vegetable oils so that there property could become solid.
Answer: N = 2.78 × 10^23 atoms
There are N = 2.78 × 10^23 atoms in 70g of Au2cl6
Completed Question:
Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits
Explanation:
Given:
Molar mass of Au2cl6 = 303.33g/mol
Mass of Au2cl6 = 70g
Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol
According to the chemical formula of Au2cl6,
1 mole of Au2cl6 contains 2 moles of Au
Number of moles of Au = 2 × 0.231mol = 0.462mole
There are 6.022 × 10^23 atoms in 1 mole of an element.
Number of Atom of gold in 0.462 mole of gold is:
N = 0.462 mol × 6.022 × 10^23 atoms/mol
N = 2.78 × 10^23 atoms
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