Answer : The mass of the water molecule is 4.5 times greater than the mass of the helium atom.
Explanation :
Assumption : The number of water molecules is equal to the number of helium atoms
Given : The mass of water = 4.5 × The mass of helium ........(1)
The mass of Water = Mass of 1 water molecule × Number of water molecule
The mass of Helium = Mass of 1 helium atom × Number of helium atom
Now these two masses expression put in the equation (1), we get
Mass of 1 water molecule × Number of water molecule = 4.5 × Mass of 1 helium atom × Number of helium atom
As per assumption, the number of water molecules is equal to the number of helium atoms. The relation between the mass of water molecule and the mass of helium atom is,
Mass of water molecule = 4.5 × Mass of helium atom
Therefore, the mass of the water molecule is 4.5 times greater than the mass of the helium atom.
Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
Answer:
The products are carbon dioxide and water
Explanation:
Step 1: Data given
Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
Step 2: The complete combustion of C3H7OH:
For the combustion of 1-propanol, we need O2.
The products of this combustion are CO2 and H2O.
C3H7OH + O2→ CO2 + H2O
On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3
C3H7OH + O2→ 3CO2 + H2O
On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.
C3H7OH + O2→ 3CO2 + 4H2O
On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).
To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.
2C3H7OH + 9O2→ 6CO2 + 8H2O
For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O
The products are carbon dioxide and water
The molarity of a solution is found to be 10 M.
Explanation:
- Molarity of a solution = <u>moles solute</u>
Litres solution
5
- Molarity of a solution = 10 M
