Answer:
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Explanation:
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Answer:
Notice that the number of atoms of
K
and
Cl
are the same on both sides, but the numbers of
O
atoms are not. There are 3
O
atoms on the the left side and 2 on the right. 3 and 2 are factors of 6, so add coefficients so that there are 6
O
atoms on both sides.
2KClO
3
(
s
)
+ heat
→
KCl(s)
+
3O
2
(
g
)
Now the
K
and
Cl
atoms are not balanced. There are 2 of each on the left and 1 of each on the right. Add a coefficient of 2 in front of
KCl
.
2KClO
3
(
s
)
+ heat
→
2KCl(s)
+
3O
2
(
g
)
The equation is now balanced with 2
K
atoms,
Answer:
The precipitated are option a and d.
Explanation:
2 LiI(aq) +Hg2(NO3)2(aq) → Hg2I2(s) ↓ + 2 LiNO3(aq)
Cation Hg2+ 2 in the presence of iodide, a precipitated is formed.
Zn(s) + 2AgNO3(aq) → 2 Ag(s) ↓ +Zn(NO3)2(aq)
Zinc starts to get rid, and some white particles also stick to it. Afterwards the solution becomes cloudy and a precipitate appears, which is the solid silver
Answer:
The resulting solution contains approximately 666 g of water.
Explanation:
In the initial solution we have:
1g salt : 8g sugar : 200g water
This means that the ratios are:
In the final solution we have:
5g salt: xg sugar: yg water
The new ratios are:
Now we can calculate the amount of sugar in the final solution:
Finally, we calculate the amount of water:
