atoms definitions and I hope this helps
Answer:
0.028kg
Explanation:
We were given 2.6 × 1023 molecules of SO 2 That means that
N = 2.6 × 1023 molecules
Use
n (number of moles) = N
NA (Avogadro's constant)
to get
n =2.6×1023 molecules
6.02 ×1023 molecules/mol = 0.432 moles
Now use
m = n × M (molar mass)
to get
m = 0.432 mol × 64.066 g/mol = 27.677 g
So
m (in kilograms) = 27.677 g
103 g/kg = 0.028 kg
The molar enthalpy of combustion in kj/mol of magnesium is 620 kj/mol, Option D is the correct answer.
<h3>What is enthalpy of Combustion ?</h3>
The energy released when a fuel is oxidized by an oxidizing agent is called enthalpy of Combustion.
It is given that
a 1.0 g sample of magnesium is burned to form MgO. in doing so, 25.5 kj of energy are released.
Molecular weight of Magnesium = 24.35g
24.35 g makes 1 mole of Mg
1g = 1/24.35
For 0.04 moles 25.5 kJ is released
for 1 mole 25.5 *1/.04
= 620 kj/mol
Therefore the molar enthalpy of combustion in kj/mol of magnesium is 620 kj/mol.
To know more about enthalpy of combustion
brainly.com/question/14754029
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<span>If one chooses to blow carbon dioxide gas into aqueous barium hydroxide it will reduce. This happens because the ions become removed from the solution and in turn this decreases the conductivity of it.</span>
Answer:
[O₃]= 8.84x10⁻⁷M
Explanation:
<u>The photodissociation of ozone by UV light is given by:</u>
O₃ + hν → O₂ + O (1)
<u>The first-order reaction of the equation (1) is:</u>
![rate = k [O_{3}] = - k \frac{\Delta [O_{3}]}{\Delta t}](https://tex.z-dn.net/?f=%20rate%20%3D%20k%20%5BO_%7B3%7D%5D%20%3D%20-%20k%20%5Cfrac%7B%5CDelta%20%5BO_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%20)
(2)
<em>where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration </em>
<u>We can get the following expression of the </u><u>first-order integrated law</u><u> of the reaction (1), by resolving the equation (2):</u>
![[O_{3}]_{t} = [O_{3}]_{0} \cdot e^{-kt}](https://tex.z-dn.net/?f=%20%5BO_%7B3%7D%5D_%7Bt%7D%20%3D%20%5BO_%7B3%7D%5D_%7B0%7D%20%5Ccdot%20e%5E%7B-kt%7D%20)
(3)
<em>where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration</em>
We can calculate the initial ozone concentration using equation (3):
![[O_{3}]_{t} = 5.0 \cdot 10^{-3}M \cdot e^{-(1.0\cdot 10^{-5}s^{-1}) (\frac{10d \cdot 24h \cdot 3600 s}{1d \cdot 1h})} = 8.84 \cdot 10^{-7}M](https://tex.z-dn.net/?f=%20%5BO_%7B3%7D%5D_%7Bt%7D%20%3D%205.0%20%5Ccdot%2010%5E%7B-3%7DM%20%5Ccdot%20e%5E%7B-%281.0%5Ccdot%2010%5E%7B-5%7Ds%5E%7B-1%7D%29%20%28%5Cfrac%7B10d%20%5Ccdot%2024h%20%5Ccdot%203600%20s%7D%7B1d%20%5Ccdot%201h%7D%29%7D%20%3D%208.84%20%5Ccdot%2010%5E%7B-7%7DM%20)
So, the ozone concentration after 10 days is 8.84x10⁻⁷M.
I hope it helps you!
