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3 years ago

Compare the evaporation of a contained liquid with that of an uncontained liquid.

1 answer:

8 0

Answer: all other conditions equal, the rate evaporation of a contained liquid will be slower than the rate of evaporation of an uncontained liquid.

Justification:

1) The rate of evaporation increases as the surface area of the liquid (relative to the whole content) increases. This is, the greater the surface is the faster the evaporation.

2) That is so because the higher the surface of the liquid the more the number of particles in the liquid that are in contact with the surrounding air and so the more the particles will escape from the liquid to the air (which is what evaporation is).

3) A liquid contained will take the form of the container, so part of the liquid wil remain below the surface, while an uncontained liquid will spread all over the surface and so pratically all the liquid is in contact witht the air surrounding it.

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The mass number of 3 isotopes of magnesium are 24,25,26.what is the mass number of the most abundant isotope of magnesium.

Anarel [89]

Answer:

24.4 amu or g/mole

Explanation:

24 x 0.790 = 19.0 amu

25 x 0.100 = 2.50 amu

26 x 0.110 = 2.86 amu

(Because of the 19.0, the sig figs go only to the 1/10 decimal place)

19.0 + 2.5 + 2.9 = 24.4 amu or g/mole

7 0

3 years ago

Calculate AGrxn for this equation, rounding your

Agata [3.3K]

Answer: $\Delta G_{rxn}=130.19J$

Explanation:

The balanced chemical reaction is,

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

The expression for Gibbs free energy change is,

$\Delta G_{rxn}=\sum [n\times \Delta G_(product)]-\sum [n\times \Delta G_(reactant)]$

$\Delta G_{rxn}=[(n_{CO_2}\times \Delta G_{CO_2})+(n_{CaO}\times \Delta G_{CaO})]-[(n_{CaCO_3}\times \Delta G_{CaCO_3})]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta G_{rxn}=[(1\times -394.4)+(1\times -604.17)]-[(1\times -1128.76)]$

$\Delta G_{rxn}=130.19J$

Therefore, the gibbs free energy for this reaction is, +130.19 kJ

4 0

3 years ago

The pyruvate dehydrogenase complex catalyzes the oxidative decarboxylation of pyruvate to form acetyl‑CoA. E1 , E2 , and E3 are

Nana76 [90]

It’s B Thiamine puro pyrophosphate (TPP)

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3 years ago

odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be add

Arte-miy333 [17]

Answer : The mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

Explanation :

First we have to calculate the moles of $H_2SO_4$ .

$\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}$

Given:

Molar mass of $H_2SO_4$ = 98 g/mole

Mass of $H_2SO_4$ = $6.05\times 10^3kg=6.05\times 10^6g$

Conversion used : (1 kg = 1000 g)

Now put all the given values in the above expression, we get:

$\text{Moles of }H_2SO_4=\frac{6.05\times 10^6g}{98g/mol}=6.17\times 10^4mol$

The moles of $H_2SO_4$ is, $6.17\times 10^4mol$

Now we have to calculate the moles of $Na_2CO_3$

The balanced neutralization reaction is:

$Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2CO_3$

From the balanced chemical reaction we conclude that,

As, 1 mole of $H_2SO_4$ neutralizes 1 mole of $Na_2CO_3$

So, $6.17\times 10^4mol$ of $H_2SO_4$ neutralizes

Now we have to calculate the mass of $Na_2CO_3$

$\text{ Mass of }Na_2CO_3=\text{ Moles of }Na_2CO_3\times \text{ Molar mass of }Na_2CO_3$

Molar mass of $Na_2CO_3$ = 106 g/mole

$\text{ Mass of }Na_2CO_3=(6.17\times 10^4mol)\times (106g/mole)=6.54\times 10^6g=6.54\times 10^3kg$

Thus, the mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

3 0

4 years ago

How does a barometer predict the weather?

Lunna [17]

Answer:

Barometers measure this pressure. ... Changes in the atmosphere, including changes in air pressure, affect the weather. Meteorologists use barometers to predict short-term changes in the weather. A rapid drop in atmospheric pressure means that a low-pressure system is arriving.

Explanation:

<h3>I hope this helps!</h3>

6 0

3 years ago