Cześć, nie mówię po polsku, więc zrobiłem to w Tłumaczu Google, więc przepraszam, jeśli coś brzmi śmiesznie.
chlor (VII) i tlen - ten wzór to 
, a tlenek chloru jest bezwodnikiem kwasu nadchlorowego.
węgiel i wodór - wzór na węgiel i wodór to CnH2n + 2), jest to związek organiczny, a niższa klasyfikacja jest taka, że jest to również węglowodór aromatyczny.
Mam nadzieję, że to pomoże, błogosławionego i cudownego dnia! :-)
-Cutiepatutie
The resulting pressure of the gas after decreasing the initial volume from 2 L to 1 L is 3 atm.
<h3>What is
Boyle's Law?</h3>
According to the Boyle's Law at constant temperature, pressure of the gas is inversely proportional to the volume of that gas.
For the given question we use the below equation is:
P₁V₁ = P₂V₂, where
P₁ = initial pressure of gas = 1.5 atm
V₁ = initial volume of gas = 2 L
P₂ = final pressure of gas = ?
V₂ = final volume of gas = 1 L
On putting all these values on the above equation, we get
P₂ = (1.5atm)(2L) / (1L) = 3 atm
Hence required pressure of the gas is 3 atm.
To know more about Boyle's Law, visit the below link:
brainly.com/question/469270
Answer:
They are more stable than alkanes
Explanation:
- <em><u>Alkenes</u></em><em><u> are a type of unsaturated hydrocarbons </u></em>which means they have a<u> double bond</u> in their structure, or lack maximum number of hydrogen atoms on each carbon.
- Alkenes have a general formula of CnH2n. They are called <u>unsaturated hydrocarbons</u> since they have a double bond. They are therefore less stable compared to alkanes and also are readily reactive.
Answer:
ΔS = +541.3Jmol⁻¹K⁻¹
Explanation:
Given parameters:
Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹
Standard Entropy of C = 5.7Jmol⁻¹K⁻¹
Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹
Standard Entropy of CO = 198Jmol⁻¹K⁻¹
To find the entropy change of the reaction, we first write a balanced reaction equation:
Fe₂O₃ + 3C → 2Fe + 3CO
To calculate the entropy change of the reaction we simply use the equation below:
ΔS = ∑S
- ∑S
Therefore:
ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1
ΔS = +541.3Jmol⁻¹K⁻¹
<span>6.38x10^-2 moles
First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
2 L / 24.465 L/mol = 0.081749438 mol
Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
0.081749438 mol * 0.78 = 0.063764562 mol.
Rounding to 3 significant figures gives 6.38x10^-2 moles</span>
