The question is incomplete, the complete question is;
AgNO3(aq)+NaCl(aq)→AgCl(s)⏐↓+NaNO3(aq)
1- If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?
Answer:
1.8 g
Explanation:
Given the equation;
AgNO3(aq)+NaCl(aq)→AgCl(s)↓+NaNO3(aq)
We can see that the reaction is 1:1 Hence, 1 mole of sodium chloride yielded 1 mole of the precipitate(AgCl).
If this is so,
Number of moles of precipitate formed = 4.52g/143.32 g/mol
Number of moles of precipitate formed = 0.0315 moles
Hence, 0.0315 moles of precipitate was formed by 0.0315 moles of NaCl
Therefore;
Mass of NaCl reacted = 0.0315 moles * 58.5 g/mol = 1.8 g
Answer:
Ghethan,Haylei,Liam
Explanation:
Liam
Haylei
Ghethan
I think that's the answer they're going for.^
Answer:-
If an element has 3 or more valence electrons then the element belongs to p block of the periodic table. For p block elements the group number is found out by adding the number of electrons in valence shell + 10.
The reason for this is because in the middle of the periodic table there are the d block elements ranging from group 3 to group 12 a total of 10 groups in between elements having 2 valence electrons and 3 valence electrons
Hello, let me break this down for you! :)
A is your energy of your reactants, B is your activation energy, so the
total value could be found from B-Y. C is the value of your energy of
your products. X is your delta H value. So in this, your products are
higher energy than the reactants, so H is + and e2-E1 is positive so E2
>E1
Therefore your answers would just be <span>E2 > E1 and
</span><span>Hrxn<span> is positive.
Hope this helps! If you have any other questions or need further explanation just let me know! :)
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