Oxygen has a strong double bond which has more stability than the single co-ordinate bond in ozone, therefore more energy is required to break the O2 bonding than ozone, so the ozone molecule is more reactive than oxygen gas. ... The oxygen free radical contains two unpaired electrons in its valence shell.
Answer:
9.91 mL
Explanation:
Using the combined gas law equation as follows;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (torr)
P2 = final pressure (torr)
V1 = initial volume (mL)
V2 = final volume (mL)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
V1 = 15.0mL
V2 = ?
P1 = 760 torr
P2 = 1252 torr
T1 = 10°C = 10 + 273 = 283K
T2 = 35°C = 35 + 273 = 308K
Using P1V1/T1 = P2V2/T2
760 × 15/283 = 1252 × V2/308
11400/283 = 1252V2/308
Cross multiply
11400 × 308 = 283 × 1252V2
3511200 = 354316V2
V2 = 3511200 ÷ 354316
V2 = 9.91 mL
Answer:
The process where substance react with oxygen is called combustion.
Explanation:
When substance react with oxygen combustion is occur. The substance which burned is called fuel and in this process large amount of heat is released to the surrounding. It is exothermic process.
For example:
4Li + O₂ → 2Li₂O
2Mg + O₂ → 2MgO
S + O₂ → SO₂
The product which is formed as a result of combustion reaction are called oxides.
In given examples we can see that lithium, magnesium and sulfur react with oxygen and product formed is oxides of respective elements such as lithium oxide ( Li₂O), magnesium oxide (MgO) and sulfur oxide ( SO₂ ).
Answer:
713.51 N/m
Explanation:
Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.
From hook's law,
F = ke ...........................Equation 1
Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.
Make k the subject of the equation,
k = F/e ............................ Equation 2
Given: F = 264 N, e = 0.37 m.
Substitute into equation 2
k = 264/0.37
k = 713.51 N/m
Hence the spring constant of the bow = 713.51 N/m
