<h2>Answer:</h2>
5.65moles
<h2>Explanations:</h2>
The formula for calculating the number of moles the compound contain is given as:
Given the following parameters
Mass of Ag = 700grams
Determine the molar mass of AgO
Molar mass = 107.87 + 16
Molar mass = 123.87g/mol
Determine the moles of AgO
Hence the moles of AgO present is 5.65moles
Answer:
The boiling point of sample X and sample Y are exactly the same.
Explanation:
The difference between sample X and sample Y is that they occupy different volumes. However, they both contain pure water. Remember that pure water has uniform composition irrespective of its volume.
Volume does not affect the boiling point as long as the volume is small enough not to give rise to significant pressure changes in the liquid.
The boiling point of a liquid is the temperature at which the pressure exerted by the surroundings upon a liquid is equaled by the pressure exerted by the vapour of the liquid; under this condition, addition of heat results in the transformation of the liquid into its vapour without raising the temperature.
It can be clearly seen from the above that the volume of a solution of pure water does not affect its boiling point hence sample X and sample Y will have the same boiling point.
Carbon atomic number ⇒ 6
Carbon mass number ⇒ 12.
Carbon atomic number - Carbon mass number = number of neutrons.
12 - 6 = 6 neutrons.
Proton charge ⇒ +1
The total charge of the nucleus of a carbon atom ⇒⇒⇒ +6.
So the naswer is (3) +6
Answer:
12.32 L.
Explanation:
The following data were obtained from the question:
Mass of CH4 = 8.80 g
Volume of CH4 =?
Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:
Mass of CH4 = 8.80 g
Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol
Mole of CH4 =?
Mole = mass/Molar mass
Mole of CH4 = 8.80 / 16
Mole of CH4 = 0.55 mole.
Finally, we shall determine the volume of the gas at stp as illustrated below:
1 mole of a gas occupies 22.4 L at stp.
Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.
Thus, 8.80 g of CH4 occupies 12.32 L at STP.
