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LenaWriter [7]
3 years ago
6

Does temperature change during a chemical reaction?​

Chemistry
2 answers:
Serga [27]3 years ago
7 0

Answer: Yes

Explanation: Temperature change is a type of chemical reaction, caused by the effect the two reactants have on each other when mixed.

Anarel [89]3 years ago
4 0

Answer: yes

Explanation:

Not all the time, but temperature change can happen during a chemical reaction.

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List the following elements in order from lowest boiling point to highest
-BARSIC- [3]

Answer:

Aluminum lodide - 680 degrees

fructose- 824 degrees

potassium bromide- 2,615 degrees

calcium bromide 3, 515 degrees

lowest boiling point is 680 and highest is 3,515.

hope this helps:)

7 0
2 years ago
What mass of a 0.583 molar solution of iron(III) nitrate is needed to obtain a) 0.0200 moles of iron(III) nitrate, b) 0.0500 mol
geniusboy [140]

Answer:

adfgbhnjm

Explanation:

34rg

8 0
3 years ago
400cm³ of dry air was passed over heated copper powder in a hard glass tube until no further change. What volume of remained at
Delicious77 [7]

Explanation:

I think there' will be a decrease in volume of the air.

3 0
2 years ago
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A company sells a natural medicine with plant ingredients to help treat dementia, a disease in which people have a hard time rem
sammy [17]
I think it is the second one. hope it helps! :)
5 0
3 years ago
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Fructose-1-P is hydrolyzed according to: Fructose-1-P + H2O → Fructose + Pi If a 0.2 M aqueous solution of Froctose-1-P is allow
aalyn [17]

Answer:

\Delta G^{\circ}=-15902 J/mol

Explanation:

In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):

\Delta G^{\circ}=-R*T*ln(K_{eq})

Being Keq:

K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}

Initial conditions:

[Fructose-1-P]=0.2M

[Fructose]=0M

[Pi]=0M

Equilibrium conditions:

[Fructose-1-P]=6.52*10^{-5}M

[Fructose]=0.2M-6.52*10^{-5}M

[Pi]=0.2M-6.52*10^{-5}M

K_{eq}=\frac{(0.2M-6.52*10^{-5}M)*(0.2M-6.52*10^{-5}M)}{6.52*10^{-5}M}

K_{eq}=613.1

Free-energy for T=298K (standard):

\Delta G^{\circ}=-8.314\frac{J}{mol*K}*298K*ln(613.1)

\Delta G^{\circ}=-15902 J/mol

3 0
3 years ago
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