Fluorine(F)
Explanation:-
- Fluorine has Z=9
- +2 charge means 2 electrons donated (7left only)
Lets look at Electronic configuration
A hockey stick is in contact with a 165-g puck for 22.4 ms; during this time, the force on the puck is given approximately by f(
t)=a+bt+ct2, where a =
1 answer:
I've found the complete form of this problem from another website which is shown in the attached picture. The equation is:
F(t) = -25 + (1.23×10⁵)t - (5.58×10⁶)t²
a.) For the first question, let's substitute t = 22.4×10⁻³ seconds to the formula.
F = -25 + (1.23×10⁵)(22.4×10⁻³ s) - (5.58×10⁶)(22.4×10⁻³ s)²
F = -69.62 N
From Newton's second law, F = ma.
-69.62 = (165 g)(1 kg/1000 g)(a)
Solving for a,
a = -421.94 m/s² = Δv/Δt = (0 - v)/(22.4×10⁻³ - 0)
Solving for v,
<em>v = 9.45 m/s</em>
b.) To solve for the distance, the formula is:
d = v₀t + 1/2(a)(t²)
Let's use the absolute value of a because distance is always positive.
d = 0(22.4×10⁻³) + 1/2(421.94)(22.4×10⁻³)²
<em>d = 0.106 m</em>
F(t) = -25 + (1.23×10⁵)t - (5.58×10⁶)t²
a.) For the first question, let's substitute t = 22.4×10⁻³ seconds to the formula.
F = -25 + (1.23×10⁵)(22.4×10⁻³ s) - (5.58×10⁶)(22.4×10⁻³ s)²
F = -69.62 N
From Newton's second law, F = ma.
-69.62 = (165 g)(1 kg/1000 g)(a)
Solving for a,
a = -421.94 m/s² = Δv/Δt = (0 - v)/(22.4×10⁻³ - 0)
Solving for v,
<em>v = 9.45 m/s</em>
b.) To solve for the distance, the formula is:
d = v₀t + 1/2(a)(t²)
Let's use the absolute value of a because distance is always positive.
d = 0(22.4×10⁻³) + 1/2(421.94)(22.4×10⁻³)²
<em>d = 0.106 m</em>
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The scheme is shown below, the steps involved are as follow,
Step one: Reduction:
The carbonyl group of given compound on reduction using Wolf Kishner reagent converts the carbonyl group into -CH₂- group.
Step two: Epoxidation:
The double bond present in starting compound when treated with m-CPBA (<span>meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.
Step three: Reduction:
The epoxide is reduced to alcohol on treatment with Lithium Aluminium Hydride (LiAlH</span>₄)<span> followed by hydrolysis.
Step four: Oxidation:
The hydroxyl group (alcohol) is oxidized to carbonyl (ketonic group) using oxidizing agent Chromic acid (H</span>₂CrO₄).
Step one: Reduction:
The carbonyl group of given compound on reduction using Wolf Kishner reagent converts the carbonyl group into -CH₂- group.
Step two: Epoxidation:
The double bond present in starting compound when treated with m-CPBA (<span>meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.
Step three: Reduction:
The epoxide is reduced to alcohol on treatment with Lithium Aluminium Hydride (LiAlH</span>₄)<span> followed by hydrolysis.
Step four: Oxidation:
The hydroxyl group (alcohol) is oxidized to carbonyl (ketonic group) using oxidizing agent Chromic acid (H</span>₂CrO₄).
Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.
Explanation : Given,
Moles of = 2.0 moles
Molar mass of = 32 g/mole
Now we have to calculate the moles of
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react to give 3 mole of
So, 2.0 moles of react to give
moles of
Now we have to calculate the mass of
Therefore, the mass of oxygen gas produced is, 96 grams.
Now we have to determine the pressure exerted by the gas against the container walls.
Using ideal gas equation:
where,
P = pressure of oxygen gas = ?
V = volume of oxygen gas
T = temperature of oxygen gas =
R = gas constant = 0.0821 L.atm/mole.K
w = mass of oxygen gas
= density of oxygen gas = 1.429 g/L
M = molar mass of oxygen gas = 32 g/mole
Now put all the given values in the ideal gas equation, we get:
Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.