Pseudoscience refers to belief, theories and practices that claims to be scientific but which have no basis in scientific facts.
Pseudoscience has many traits two of which are:
1. It is based on dogma and uncritical belief. Counter evidence might draw hostility from pseudo scientists.
2. There is usually suppression or distortion of data. Pseudo scientists typically perform selective data gathering, in which they only gather and consider favorable data that support their claims. <span />
Answer:
k = 4,92x10⁻³
Explanation:
For the reaction:
AB₂C (g) ⇄ B₂(g) + AC(g)
The equilibrium constant, k is defined as:
![k = \frac{[B_{2}][AC]}{[AB_2C]}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B%5BB_%7B2%7D%5D%5BAC%5D%7D%7B%5BAB_2C%5D%7D)
<em>(1)</em>
Molar concentration of the species are:
[AB₂C]: 0,0840mol / 5L = <em>0,0168M</em>
[B₂]: 0,0350mol / 5L = <em>0,0070M</em>
[AC]: 0,0590mol / 5L = <em>0,0118M</em>
Replacing this values in (1):
![k = \frac{[0,0070][0,0118]}{[0,0168]}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B%5B0%2C0070%5D%5B0%2C0118%5D%7D%7B%5B0%2C0168%5D%7D)
<em>k = 4,92x10⁻³</em>
I hope it helps!
Answer:
A condensation reaction
Explanation:
it is also called a dehydration reaction
Answer:
C₆H₆O₃
Explanation:
Calculation sequence:
% => grams => moles => reduce => empirical Ratio
Molecular multiple = Molecular Mass / Empirical Mass
C: => 57.1% => 57.1 g => 57.1/12 = 4.7583
H: => 4.8% => 4.8 g => 4.8/1 = 4.8000
O: => <u>38.1% => 38.1 g </u>=> 38.1/16 = 2.3813
TTL => 100% 100 g
Reduced Mole values =>
C : H : O => 4.7583/2.3813 : 4.8000/2.3813 : 2.3813/2.3813 => 2 : 2 : 1
∴ empirical formula => C₂H₂O
empirical formula weight => 2C + 2H + 1O = [2(12) + 2(1) + 1(16)] amu = 42 amu
molecular formula weight (given in problem) = 126 g/mole
The molecular formula is a whole number multiple of the empirical formula.
molecular multiple = 126 amu / 42 amu = 3
∴ molecular formula => (C₂H₂O)₃ => C₆H₆O₃
Answer:
Energy lost is 7.63×10⁻²⁰J
Explanation:
Hello,
I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5
E = hc/λ(1/n₂² - 1/n₁²)
n₁ = 15
n₂ = 5
hc/λ = 2.18×10⁻¹⁸J (according to the data)
E = 2.18×10⁻¹⁸ (1/n₂² - 1/n₁²)
E = 2.18×10⁻¹⁸ (1/15² - 1/5²)
E = 2.18×10⁻¹⁸ ×(-0.035)
E = -7.63×10⁻²⁰J
The energy lost is 7.63×10⁻²⁰J
Note : energy is lost / given off when the excited electron jumps from a higher energy level to a lower energy level
