4 carbon electrons. Hope this helps have a nice day
Answer:
Option A. 1.14L
Explanation:
Data obtained from the question include:
V1 = 1.20 L
P1 = at stp = 760torr
T1 = at stp = 273K
P2 = 900torr
T2 = 35°C = 35 + 273 = 308K
V2 =?
Using P1V1/T1 = P2V2/T2, we can obtain the new volume as follows:
P1V1/T1 = P2V2/T2
(760 x 1.2)/273 = (900 x V2) /308
Cross multiply to express in linear form
273 x 900 x V2 = 760 x 1.2 x 308
Divide both side by 273 x 900
V2 = (760 x 1.2 x 308) /(273 x 900)
V2 = 1.14L
Answer:
42.65g
Explanation:
Given parameters:
Mass of K = 4g
Unknown: Mass of KCl
Solution:
Complete equation of the reaction:
2K + Cl₂ → 2KCl
To solve this problem, we know that the reactant in short supply is potassium K and this dictates the amount of products that would be formed. The chlorine gas is in excess and we can't use it to determine the amount of product that would form.
Now, we work from the known to the unknown. Since we know the mass of K given in the reaction, we can simply find the molar relationship between the reacting potassium and the product. We simply convert the mass to mole and compare to the product. From there we can find the mass of KCl that would be produced.
Calculating number of moles of K
Number of moles = 
Number of moles of K =
= 0.103mol
From the given reaction equation:
2 moles of K will produce 2 moles of KCl
Therefore 0.103mol of K will produce 0.103mol of KCl
To find the mass of KCl produced,
Mass of KCl = number of moles of KCl x molar mass
Molar mass of KCl = 39 + 35.5 = 74.5gmol⁻¹
Mass of KCl = 0.103 x 74.5 = 42.65g
<span>In determining significant figure, always remember the rules:
Zeros place in the beginning and end of a non-zero numbers are not significant
Zeros place in the middle of a non-zero numbers are considered significant.
Therefore, 340 438 grams has 6
significant figures.</span>