Answer:
7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.
Explanation:
First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.
According to the given question we have to prepare 0.100 M solution
1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4
1 ml of solution contain 14.208÷1000= 0.014 gram
0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.
So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.
That’s because when a charged chemical species composes of two or more atoms (covalently bonded), they act as a single unit. the term radicals refers to free radicals that are with an unpaired electron and because not all of its electrons are found in pairs
Answer:
<em>Hi Todoroki here!!! </em>
Explanation:
Chlorine has the electron configuration [Ne]3s 2 3p 5, with the seven electrons in the third and outermost shell acting as its valence electrons. Like all halogens, it is thus one electron short of a full octet, and is hence a strong oxidising agent, reacting with many elements in order to complete its outer shell.
<em>Your welcome!!</em>
My best try would be d an element
<span>Cations mix with anions, so you know NH4+ won't mix with K+ and SO4(2-) won't miix with F-. For the reason that NH4+ and F- together have single charges, they'll mix in a 1:1 ratio, NH4F. There's two charges on SO4(2-), so it'll need two K+ to mix with, K2SO4.</span>
