Answer:
12.32 L.
Explanation:
The following data were obtained from the question:
Mass of CH4 = 8.80 g
Volume of CH4 =?
Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:
Mass of CH4 = 8.80 g
Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol
Mole of CH4 =?
Mole = mass/Molar mass
Mole of CH4 = 8.80 / 16
Mole of CH4 = 0.55 mole.
Finally, we shall determine the volume of the gas at stp as illustrated below:
1 mole of a gas occupies 22.4 L at stp.
Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.
Thus, 8.80 g of CH4 occupies 12.32 L at STP.
Answer:
Explanation:
Both motion of particles and temperature increase
Answer:
Explanation:
Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)
Explanation;
According to Arrhenius equation:
i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)
Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K
T2 = 25 oC = (25 + 273) K = 298 K
i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)
i.e. ln(k2/0.000717) = -2.54738
i.e. k2/0.000717 = 
= 0.078286
Therefore, the required constant (k2) = 0.078286 * 0.000717 = 
Answer:
The three-step synthesis of trans-2-pentene from acetylene is as follows.
<u>Step -1:</u> Formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkanes.
<u>Step -2:</u> Formation terminal alkyne to nonterminal alkynes.
<u>Step -3:</u> Formation of trans-pent - 2-pent-ene by reduction.
Explanation:
Synthesis of trans-pent-2-yne from ethyne takes place is mainly a three step synthesis which involves formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkane. Second step involves the further alkylation of terminal alkynes to higher order nonterminal alkynes and the third step involves the formation of trans-2-ene by dissolving reduction method.
The chemical reaction of each step of chemical reactions is as follows.
Denitrification is part of the nitrogen cycle