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polet [3.4K]
3 years ago
11

Which energy transformation occurs when an

Chemistry
1 answer:
DerKrebs [107]3 years ago
8 0

I believe that it is 1. if not, 2. but most likely, in my opinion, 1
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Select the correct answer.
kirza4 [7]

Answer: A. It indicates the direction in which the reaction occurs.

4 0
1 year ago
If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L
Tom [10]

Answer:

Approximately 2.53\; \rm L (rounded to three significant figures) assuming that {\rm HCl}\, (aq) is in excess.

Explanation:

When {\rm HCl} \, (aq) and {\rm AgNO_3}\, (aq) precipitate, {\rm AgCl} \, (s) (the said precipitate) and \rm HNO_3\, (aq) are produced:

{\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq) (verify that this equation is indeed balanced.)

Look up the relative atomic mass of \rm Ag and \rm Cl on a modern periodic table:

  • \rm Ag: 107.868.
  • \rm Cl: 35.45.

Calculate the formula mass of the precipitate, \rm AgCl:

\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm  g\cdot mol^{-1}\end{aligned}.

Calculate the number of moles of \rm AgCl formula units in 45.00\; \rm g of this compound:

\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}.

Notice that in the balanced equation for this reaction, the coefficients of {\rm AgNO_3} \, (aq) and {\rm AgCl}\, (s) are both one.

In other words, if {\rm HCl}\, (aq) (the other reactant) is in excess, it would take exactly 1\; \rm mol of {\rm AgNO_3} \, (aq)\! formula units to produce 1\; \rm mol \! of {\rm AgCl}\, (s)\! formula units.

Hence, it would take 0.313987\; \rm mol of {\rm AgNO_3} \, (aq)\! formula units to produce 0.313987\; \rm mol\! of {\rm AgCl}\, (s)\! formula units.

Calculate the volume of the {\rm AgNO_3} \, (aq)\! solution given that the concentration of the solution is 0.124\; \rm mol \cdot L^{-1}:

\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}.

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of {\rm AgNO_3} \, (aq)\!.)

In other words, approximately 2.53\; \rm L of that {\rm AgNO_3} \, (aq)\! solution would be required.

4 0
2 years ago
Srg unf msmt. I’m so lost
Gala2k [10]

i cannot see what the question is?

4 0
3 years ago
Placer deposits form when ____.
Oxana [17]
Think it’s B or C because they make more sense
8 0
2 years ago
Read 2 more answers
Identify how many fluorine atoms are in one molecule of sulfur hexafluoride. How do you know this? Read More >>
GaryK [48]
<span>hexa stands for 6, so this molecule has 1 sulfur atom, surrounded by 6 fluorine atoms.</span>
6 0
3 years ago
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