Hey there!:
Detailed solution is shown below ask if any doubt :
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 = 
The question is incomplete.
You need two additional data:
1) the original volume
2) what solution you added to change the volume.
This is a molarity problem, so remember molarity definition and formula:
M = n / V in liters: number of moles per liter of solution
To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).
The original solution was:
V= 1 ml
M = 0.2 M
Using the formula for molarity, M = n / V
n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles
For the final solution:
n = 0.0002 moles
M = 0.04
From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l
Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml. This would be the answer for the hypothetical problem that I assumed for you.
I hope this gives you all the cues you need to answer similar problems about molarity.
Answer: Metals form cations.
The alkali metals (the IA elements) lose a single electron to form a cation with a 1+ charge.
The alkaline earth metals (IIA elements) lose two electrons to form a 2+ cation.
Aluminum, a member of the IIIA family, loses three electrons to form a 3+ cation.
Therefore, metals in the s and p block of the periodic table have 1, 2 or 3 electrons in their outermost orbit (or valence shell). Now to gain a stable octet metals lose either 1, 2 or 3 electrons from the valence shell thus forming cation with +1, +2 or +3 charge.
X it by the molar mass of tungsten
