Answer:
c) Hexagonal close packed
Explanation:
The fact that a material is crystalline, amorphous or semicrystalline defines some of its properties. For example, if a material had the option of being crystalline or amorphous, it would be observed that in the crystalline state the density of the material would be greater in the amorphous state. This is because because of the order that the atoms follow in the crystalline material, it is possible to place more atoms in the same physical volume. When there are more atoms there is more mass in the same volume so the density is higher than when the material is amorphous.
The different crystalline structures are the following:
Simple cubic structures: the unit cell is a common edge cube, with a defined network point in each of its vertices.
Cubic structure centered on the body: the unit cell is a cube with a common edge. It has a defined network point in each of its vertices and a defined network point in the geometric center of the cube. INCLUDES a)
Cubic structure centered on the face: the unit cell is a cube with a common edge. It has a defined network point in each of its vertices and a defined network point in the geometric center of each of its faces. INCLUDES b) and d)
The volume of water vapour would be produced at 19°C and 780 torr is 548.5mL.
If 400 ml of CO2 is produced at 30°C at 740 torr, then number of moles can be calculated as:
By using ideal gas equation:
P1V1 = N1R1T1
P1 = pressure = 740torr
V1 = 400 ml = volume of CO2
R = Gas constant = 8.314
T = 273+30 = 303 k
740×400 = N1×8.314×303
N1 = (740×400) /(8.314×303) =117.5.
Chemical equation
C2H6 ---- 2CO2 + 3H2O.
As we noticed from the equation that
2 moles of CO2 = 3 moles of H2O
1 moles of CO2 × 1 moles of H2O
Then N2 = 117.5 moles of CO2 = 3/2 × 117.5 moles of H2O
By using ideal gas equation:
P2V2 = N2RT2
V2 = 3/2 × 117.5 × 8.314 × 292/ 780
= 548.5ml.
Thus, we found that the volume of water vapour would be produced at 19°C and 780 torr is 548.5mL.
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Answer:
Applications of Chemistry:
Chemistry plays an important and useful role towards the development and growth of a number of industries. This includes industries like glass, cement, paper, textile, leather, dye etc. We also see huge applications of chemistry in industries like paints, pigments, petroleum, sugar, plastics, Pharmaceuticals.
Taking into account the scientific notation, the result of the subtraction is -4.20689×10⁻².
<h3>Scientific notation</h3>
Scientific notation is a quick way to represent a number using powers of base ten, where the numbers are written as a product:
a×10ⁿ
where:
- a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a non-integer number.
- n is an integer, which is called an exponent or an order of magnitude. Represents the number of times the comma is shifted. It is always an integer, positive if it is shifted to the left, negative if it is shifted to the right.
<h3>Subtraction in scientific notation</h3>
When the numbers to be added do not have the same base 10 exponent, the base 10 power with the highest exponent must be found. In this case, the highest exponent is -2.
Then all the values are expressed as a function of the base 10 exponent with the highest exponent. In this case: 3.11×10⁻⁵= 0.00311×10⁻²
Taking the quantities to the same exponent, all you have to do is subtract what was previously called the number "a". In this case:
0.00311×10⁻² - 4.21×10⁻²= (0.00311 - 4.21)×10⁻²= -4.20689×10⁻²
Finally, the result of the subtraction is -4.20689×10⁻².
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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!
