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LiRa [457]
3 years ago
15

Complete the equation: monosaccharide monosaccharide → ______ water

Chemistry
1 answer:
Lady_Fox [76]3 years ago
5 0
If I understand you correctly, then this is the equation for the breakdown of monosaccharide.
monosaccharide --> Carbon dioxide + water
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Does water added to a flask containing chlorine gas and solid sodium react in the chemical reaction to make sodium chloride
Jet001 [13]

Answer:

Explanation:

Chemically, we can have a reaction between chlorine gas and solid sodium

This reaction is actually explosive and would produce fine powder of sodium chloride

We have the reaction as follows:

2Na_{(s)}+Cl_{2(g)}\rightarrow2NaCl_{(s)}

Essentally, what we can deduce from here is that we do not need to add water to the flask. Except for the reason that we would want the sodium chloride solid in the solution form, there is absolutely no reason to add water to the flask as the reaction would proceed normally

5 0
1 year ago
Helppppp pleaseeee xxxxxx
kogti [31]

Answer:

2812.6 g of H₂SO₄

Explanation:

From the question given above, the following data were obtained:

Mole of H₂SO₄ = 28.7 moles

Mass of H₂SO₄ =?

Next, we shall determine the molar mass of H₂SO₄. This can be obtained as follow:

Molar mass of H₂SO₄ = (1×2) + 32 + (16×4)

= 2 + 32 + 64

= 98 g/mol

Finally, we shall determine the mass of H₂SO₄. This can be obtained as follow:

Mole of H₂SO₄ = 28.7 moles

Molar mass of H₂SO₄ =

Mass of H₂SO₄ =?

Mole = mass / Molar mass

28.7 = Mass of H₂SO₄ / 98

Cross multiply

Mass of H₂SO₄ = 28.7 × 98

Mass of H₂SO₄ = 2812.6 g

Thus, 28.7 mole of H₂SO₄ is equivalent to 2812.6 g of H₂SO₄

3 0
3 years ago
Draw the products of the complete hydrolysis of an acetal. Draw all products of the reaction.
jolli1 [7]

Answer: the product is ketone or aldehyde

Explanation:

The first step is the conversion of acetal to hemiacetal in the presence of H3O+/ ROH, and then the final conversion of hemiacetal to ketone/aldehyde using

H3O+/ ROH...

Attached is the structural conversion

8 0
3 years ago
15 POINTS PLEASE HELP What volume of water must be added to 35mL of 2.6m KCl to reduce its concentration to 1.2m? Please explain
BartSMP [9]
First, find the volume the solution needs to be diluted to in order to have the desired molarity:
You have to use the equation M₁V₁=M₂V₂ when ever dealing with dilutions.

M₁=the starting concentration of the solution (in this case 2.6M)
V₁=the starting volume of the solution (in this case 0.035L)
M₂=the concentration we want to dilute to (in this case 1.2M)
V₂=the volume of solution needed for the dilution (not given)

Explaining the reasoning behind the above equation:
MV=moles of solute (in this case KCl) because molarity is the moles of solute per Liter of solution so by multiplying the molarity by the volume you are left with the moles of solute.  The moles of solute is a constant since by adding solvent (in this case water) the amount of solute does not change.  That means that M₁V₁=moles of solute=M₂V₂ and that relationship will always be true in any dilution.

Solving for the above equation:
V₂=M₁V₁/M₂
V₂=(2.6M×0.035L)/1.2M
V₂=0.0758 L
That means that the solution needs to be diluted to 75.8mL to have a final concentration of 1.2M.

 Second, Finding the amount of water needed to be added:
Since we know that the volume of the solution was originally 35mL and needed to be diluted to 75.8mL to reach the desired molarity, to find the amount of solvent needed to be added all you do is V₂-V₁ since the difference in the starting volume and final volume is equal to the volume of solvent added.
75.8mL-35mL=40.8mL
40.8mL of water needs to be added

I hope this helps.  Let me know if anything is unclear.
Good luck on your quiz!
5 0
3 years ago
Which elements, when they have to, can have more than
inn [45]
Elements in the third row can break the octet rule
6 0
3 years ago
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