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3 years ago

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Does evaporation happen before of after condensation?

2 answers:

Naya [18.7K]3 years ago

8 0

Answer:

Does evaporation happen before of after condensation?

Explanation:

Condensation, the opposite of evaporation, occurs when saturated air is cooled below the dew point (the temperature to which air must be cooled at a constant pressure for it to become fully saturated with water), such as on the outside of a glass of ice water.

Or

Heat (energy) is necessary for evaporation to occur. Condensation, the opposite of evaporation, occurs when saturated air is cooled below the dew point (the temperature to which air must be cooled at a constant pressure for it to become fully saturated with water), such as on the outside of a glass of ice water.

rodikova [14]3 years ago

7 0

Answer:

before

Explanation:

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ZILLDIFFEQMODAP11 3.1.021. My Notes Ask Your Teacher A tank contains 150 liters of fluid in which 10 grams of salt is dissolved.

SSSSS [86.1K]

Answer:

Therefore, the number of grams of salt in the tank at time t is $A(t) = 150-140 e^{-\frac{t}{30} }$

Explanation:

Given:

Tank A contain $V_{1} = 150$ lit

Rate $\alpha = 5 \frac{L}{min}$

Dissolved salt $A = 10$ gm

Salt pumped in one minute is $4 \frac{L}{min}$

Salt pumped out is $\frac{5L}{150L} = \frac{1}{30}$ of initial amount added salt.

To find $A(t)$

$\frac{dA}{dt} = Rate _{in} - Rate _{out}$

$A' = 5 - \frac{A}{30}$

$A' + \frac{A}{30} = 5$

Solving above equation,

$I .F = e^{\int\limits {p} \, dt }$

$y = e^{\int\limits {\frac{1}{30} } \, dt }$

$y = e^{\frac{t}{30} }$

$(Ae^{\frac{t}{30} } )' = 5 e^{\frac{t}{30} } + c$

Integrating on both side,

$Ae^{\frac{t}{30} } = 5 \times 30 e^{\frac{t}{30} } +c$

Add $e^{-\frac{t}{30} }$ on above equation,

$A = 150 + ce^{-\frac{t}{30} }$

Here given in question, $A(t=0) = 10$

$10 =150 +c$

$c = -140$

Put value of constant in above equation, and find the number of grams of salt in the tank at time t.

$A(t) = 150-140 e^{-\frac{t}{30} }$

Therefore, the number of grams of salt in the tank at time t is $A(t) = 150-140 e^{-\frac{t}{30} }$

7 0

3 years ago

Find the molarity: 9.82 grams of lead (IV) nitrate are dissolved to make 465 mL of solution.

solmaris [256]

Molarity is moles per liter. Therefore you must covert grams->moles by dividing 9.82 by the molar mass of lead IV nitrate. Then convert mL->L. Once you have those two numbers you can plug them into the formula Molarity = mol/L.

3 0

3 years ago

What is the final volume (L) of a 1.00 L system at 315 K and 1.10 atm if STP conditions are established?

LenaWriter [7]

Answer:

0.95L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 1L

T1 (initial temperature) = 315K

P1 (initial pressure) = 1.10 atm

T2 (final temperature) = stp = 273K

P2 (final pressure) = stp = 1atm

V2 (final volume) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final volume of the system can be obtained as follow:

P1V1/T1 = P2V2/T2

1.1 x 1/315 = 1 x V2/273

Cross multiply to express in linear form.

315 x V2 = 1.1 x 273

Divide both side by 315

V2 = (1.1 x 273) /315

V2 = 0.95L

Therefore, the final volume of the system if STP conditions are established is 0.95L

5 0

3 years ago

Is this a redox reaction? give evidence (many points plz answer)<br> 2Mg + CO2 → 2MgO + C

Aleonysh [2.5K]

Answer:

$\boxed{yeah}$

Explanation:

$CO_2 \: is \: reduced \: to→C \: by \: 2M_g \: (the \: reducing \: agent): while \\ 2M_g \: is \: oxydized \: to→2M_gO \: by \: CO_2 \: (the \: oxydizing \: agent)$

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3 years ago

For the reaction system, h2(g) + x2(g) ↔ 2 hx(g), kc = 24.4 at 300 k. a system made up from these components which is at equilib

Pavel [41]

Kc=24.4=[HX]∧2/[H2]×[X2] =(0.6)∧2/(0.2)×[H2]
[H2] = 0.36/(24.4×0.2) = 0.07377 mole

4 0

4 years ago

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