Answer:
60 cm³ of water
Explanation:
We'll begin by calculating the volume of the diluted solution. This can be obtained as follow:
Concentration of stock solution (C₁) = 17 M
Volume of stock solution (V₁) = 25 cm³
Concentration of diluted solution (C₂) = 5 M
Volume of diluted solution (V₂) =?
C₁V₁ = C₂V₂
17 × 25 = 5 × V₂
425 = 5 × V₂
Divide both side by 5
V₂ = 425 / 5
V₂ = 85 cm³
Thus, the volume of the diluted solution is 85 cm³
Finally, we shall determine the volume of water needed to dilute the solution. This can be obtained as follow:
Volume of stock solution (V₁) = 25 cm³
Volume of diluted solution (V₂) = 85 cm³
Volume of water =?
Volume of water = V₂ – V₁
Volume of water = 85 – 25
Volume of water = 60 cm³
Therefore, 60 cm³ of water is needed to dilute the solution.
Answer:
glycerol, pyruvate, glycerol -3-phosphate, dihydroxyacetone phosphate and glucose♡ hope this helps♡
Density of the vinegar is higher than the density of the oil.
Explanation:
Density of the vinegar is higher than the density of the oil. The consequence of this is that the oil will be the top layer in the pitcher while the vinegar is at the bottom layer in the pitcher.
When mixing oil and vinegar will not produce a mixture because the oil contains non-polar molecules while vinegar is a solution of acetic acid in water and both of them are polar molecules.
Learn more about:
liquids with different densities
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Answer:
Alkanes with more than 3 carbons can show constitutional isomerism. They can be either linear or branched structures. This is categorized as chain isomerism. Butane is the smallest alkane to show such isomerism with 2 isomers.Alkanes with more than 3 carbons can show constitutional isomerism. They can be either linear or branched structures. This is categorized as chain isomerism. Butane is the smallest alkane to show such isomerism with 2 isomers.
Explanation:
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,
