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S_A_V [24]
3 years ago
6

What is another way of describing a freezing point change?

Chemistry
2 answers:
Paladinen [302]3 years ago
7 0
Melting is equivalent to freezing: a temperature when a solid become a liquid.
andrew-mc [135]3 years ago
6 0
The girl on top is correct
You might be interested in
When PCl5 solidifies it forms PCl4+ cations and PCl6– anions. According to valence bond theory,
rewona [7]

Hey there!

No of hybrid orbitals , H = ( V +S - C + A ) / 2

Where H = no . of hybrid orbitals

V = Valence of the central atom = 5

S = No . of single valency atoms = 4

C = No . of cations = 1

A = No . of anions = 0

For PCl4 +

Plug the values we get H = ( 5+4-1+0) / 2

H =  4 ---> sp3 hybridization

sp3 hybrid orbitals are used by phosphorous in the PCl4+ cations

Answer C

Hope that helps!

3 0
4 years ago
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

7 0
3 years ago
IF UR GOOD AT CHEM PLZ ANSWER MY PREVIOUS QUESTIONS ON MY PROFILE, I NEED HELP!!
erastova [34]

Answer:

Ok, I'll try

Explanation:

7 0
3 years ago
24 grams of CH4 was added to the above reaction. Calculate the theoretical yield of CO2. A. 66 grams B. 33 grams c. 132 grams. D
Sonbull [250]

Answer:

Option A. 66 g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the mass of CH₄ that reacted and the mass of CO₂ produced from the balanced equation. This is illustrated below:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4 = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32 = 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Finally, we shall determine the theoretical yield of CO₂. this can be obtained as follow:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Therefore, 24 g of CH₄ will react to produce = (24 × 44) /16 = 66 g of CO₂.

Thus, the theoretical yield of CO₂ 66 g

8 0
3 years ago
During which of the following phases of the moon do we see the left half of the moon as lit?
xxMikexx [17]
During the 3rd quarter
7 0
3 years ago
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