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EastWind [94]
3 years ago
15

How do you describe how sugar dissolves in tea using the big idea of particles?

Chemistry
2 answers:
Alisiya [41]3 years ago
4 0
If you use big pieces of sugar, the particles will dissolve harder...As the pieces are smaller, as much they dissolve faster.
Komok [63]3 years ago
3 0
Sugar is an ionic compound and tea is most likely a polar substance. When the sugar is placed in the tea the molecules of the tea attack the sugar molecules using their negative and positive pulls. The negative sides of the tea molecules pull at the positive parts of the sugar, and vice versa. <span />
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An aqueous solution of potassium sulfate is allowed to react with an aqueous solution of calcium nitrate. identify the solid in
Maksim231197 [3]
Ans: Calcium sulfate.

K2SO4 (aq) + Ca(NO3)2 (aq) ⇒ 2KNO3 (aq) + CaSO4 (s)

6 0
3 years ago
I am a nonmetal.I am in the Oxygen family and in row 3.I have 6 valence electrons.I am yellow and have a stinky smell.Who am i
Dimas [21]

Answer:

sulfur

Explanation:

In oxygen family sulfur has yellow color and also having stinky smell. Thus given statements are about sulfur.

It is present in oxygen family.

It has six valance electrons.

Its atomic number is 16.

Its atomic weight is 32 amu.

The electronic configuration of sulfur is given below,

S₁₆ = 1s² 2s² 2p⁶ 3s² 3p⁴

We can see the valance shell is third shell and it have six electrons thus sulfur have six valance electrons. (3s² 3p⁴ )

Sulfur is used in vulcanisation process.

It is used in bleach and also as a preservative for many food.

it is used to making gun powder.

8 0
2 years ago
What is the atomic mass of copper?
kipiarov [429]

Answer:

63.546 u

Explanation:

4 0
2 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
A certain half-reaction has a standard reduction potential E0red = +0.13V . An engineer proposes using this half-reaction at the
Ivan

Answer:

a. 1.23 V

b. No maximum

Explanation:

Required:

a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?

b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

If E°cell must be at least 1.10 V (E°cell > 1.10 V),

E°red, cat - E°red, an > 1.10 V

E°red, cat - 0.13V > 1.10 V

E°red, cat > 1.23 V

The minimum standard reduction potential is 1.23 V while there is no maximum standard reduction potential.

4 0
3 years ago
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