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3 years ago

What is the affect of increasing the water's mass?how does it reflect it's temperature?

1 answer:

8 0

I assume what you're asking about is, how does the temperature changes when we increase water's mass, according the formula for heat ?
Well the formula is : $Q=m\cdot c\cdot \Delta t$ (where Q is heat, m is mass, c is specific heat and $\Delta t$ is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless, if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature

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Research is being carried out on cellulose as a source of chemicals for the production of fibers, coatings, and plastics. Cellul

Luden [163]

Answer:

+ 636 KJ

Explanation:

We want to arrive to the equation

C6H12O6(s) ---------> 6 H2CO(g) ΔH ° rxn = ?

by manipulating algebraically the first four given equations.

We notice the first one has our product H2CO(g) as a reactant. This indicates we must take the inverse of that equation. Also we need 6 mol of H2CO(g), thus it also needs to be multiplied by 6

6 CO2(g) + 6 H2O(g) ------->6H2CO(g) + 6O2(g) ΔH °comb = + 572.9 KJ/x 6

Now we want C6H12O6(s) as a reactant and it is a product in the second one, therefore lets reverse it

C6H12O6(s) -------> 6 C(s) + 6 H2(g) + 3 O2(g) ΔH ° f = + 1274.4 KJ/mol

Now if take the third equation and multiply it by six we will cancel the C(s) with the above equation

6 C(s) + 6O2(g) ---------> 6 CO2(g) ΔH ° f = - 393.5 KJ/mol x 6

Finally by multiplying the last equation by 6 and adding all the equations we will arrive at our desired one

6 H2(g) + 3 O2(g) -----------> 6H2O(g) ΔH ° f = - 285.8 KJ/mol x 6

then lets add them to get ΔH ° rxn:

6 CO2(g) + 6 H2O(g) ------->6H2CO(g) + 6O2(g) ΔH °comb = + 3437.4 KJ

+ C6H12O6(s) -------> 6 C(s) + 6 H2(g) + 3 O2(g) ΔH ° f = + 1274.4 KJ

+ 6C(s) + 6O2(g) ---------> 6 CO2(g) ΔH ° f = - 2361.0 KJ

+6 H2(g) + 3 O2(g) -----------> 6H2O(g) ΔHº f = - 1714.8 KJ

C6H12O6(s) ---------> 6 H2CO(g)

ΔH ° rxn = 3437.4 + 1274.4 - 2361.0 - 1714.8 = 636 KJ

8 0

3 years ago

The formation of a plaque and halitosis can be avoided by brushing and flossing at least :

Jet001 [13]

Answer:

at least 3 times a day

Explanation:

7 0

3 years ago

A chemist must prepare of potassium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask

maksim [4K]

The question is incomplete; the complete question is:

A chemist must prepare 800.0mL of potassium hydroxide solution with a pH of 13.00 at 25 degree C. He will do this in three steps: Fill a 800.0mL volumetric flask about halfway with distilled water. Weigh out a small amount of solid potassium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to 2 significant digits.

Answer:

4.5g (to 2 significant digits)

Explanation:

Now we must remember that KOH is a strong base, therefore it will practically dissociate completely.

To find the pH we can use the equation pH + pOH = 14.

Firstly to find the pOH:

13.00 + pOH = 14

pOH = 1.00

To find the [OH-]

Since

pOH= -log[OH^-]

[OH^-] = antilog (-pOH)

[OH^-]= antilog (-1)

[OH^-] = 0.1 molL-1

Since we've established that KOH is a strong base, we know that [OH-] = [KOH]

Also, we know that concentration = number of moles/volume

we have the concentration and the volume now so we can calculate the number of number of moles as follows:

The 800mL volume is the same as 0.8L

0.1 molL-1= number of moles/0.8L

0.08 moles = number of moles

now we can calculate the amount of solid KOH required

the molar mass of KOH = 39 + 16 +1 = 56 gmol-1

56 x 0.08 moles = 4.48g

So in 800mL of pH 13.00 KOH there is 4.5g of KOH dissolved.

3 0

3 years ago

Write the structure of the aldol condensation-dehydration product that you synthesized. Using this structure, analyze the NMR sp

STALIN [3.7K]

Answer:

Please find the solution in the attached file.

Explanation:

8 0

2 years ago

Technically speaking, in terms of chemistry and physics, steam is visible and can be seen.

Nezavi [6.7K]

A true, hope it’s right

8 0

3 years ago

Read 2 more answers