Which of the following is the most reactive?

Can u help me with this

dolphi86 [110]

Answer:

c

Explanation:

the reason is because your a homosiejenenddj

8 0

2 years ago

An exhaled air bubble underwater at 290.

Vinil7 [7]

The answer for the following problem is mentioned below.

Therefore the final volume of the gas is 52.7 ml.

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 290 kPa

Final pressure ( $P_{2}$ ) = 104 kPa

Initial volume ( $V_{1}$ ) = 18.9 ml

To find:

Final volume ( $V_{2}$ )

We know;

From the ideal gas equation;

P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of gas

n represents the no of the moles

R represents the universal gas constant

T represents the temperature of the gas

So;

P × V = constant

P ∝ $\frac{1}{V}$

From the above equation;

$\frac{P_{1} }{P_{2} } = \frac{V_{2} }{V_{1} }$

$P_{1}$ represents the initial pressure of the gas

$P_{2}$ represents the final pressure of the gas

$V_{1}$ represents the initial volume of the gas

$V_{2}$ represents the final volume of the gas

Substituting the values of the above equation;

$\frac{290}{104}$ = $\frac{V_{2} }{18.9}$

$V_{2}$ = 52.7 ml

Therefore the final volume of the gas is 52.7 ml.

6 0

3 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago

Choose all that apply. As glaciers melted at the end of the last Ice Age:

Kaylis [27]

As glaciers melted at the end of the last Ice Age worldwide sea level increased immensely and river plains were flooded. Glaciers are floating bodies of ice. Ice is frozen water. When glaciers started melting water that had been frozen for decades or centuries was released back into the ocean. That lead to an increase in sea level. This also led to river plains being drowned/flooded. Therefore, the answer is 1 and 2.

3 0

3 years ago

Read 2 more answers

PLEASE SOMEBODY EXPLAIN THIS :(((

djverab [1.8K]

Answer:

$\boxed{ \sf \: R_f \: value \: of \: sample \: 1 =0.3142}$

<h3>Explanation:</h3>

In Analytical Chemistry chromatography is widely used for the separation of samples.

In thin layer chromatography, the mixture of components are separated on the basis of their polarity.
The solvent solution(mobile phase) that we use are non polar & silica gel( TLC paper made of/stationary phase) are polar.
Consider the mixture we have taken consist of two samples having large polar difference.
Due to opposite nature of silica gel(polar) & solvent solution (non polar) the movement become easy & due to capillary action solvent solution rise to the top.
The mixture of sample we have taken, the sample have less polarity have high peak or they travel more distance than that of more polar sample when they dipped into the solution.

In the given diagram, mixture of 8 samples are separated on the basis of their polarity, the distance travelled by solvent is 35 mm, distance travelled by sample 1 is 11 mm & similarly distance travelled by sample 2,3,4,5,6,7 are 15,31,4,22,25,33 in mm respectively.

Rf Value: Rf value is retention factor which tells about relative absorption of each sample & range of Rf value is 0-1.

Formula to calculate Rf value is

$\sf R_f \: value = \frac{distance \: moved \: by \: sample}{distance \: moved \: by \: solvent}$

Now, solving for Rf value of sample 1

Given:

Distance moved by sample 1 = 11 mm

Distance movedby solvent = 35 mm

To find:

Rf value of sample 1 = ?

Solution:

Substituting the given data in above formula,

$\small \sf R_f \: value = \frac{distance \: moved \: by \: sample \: 1}{distance \: moved \: by \: solvent} \\ \small \sf R_f \: value = \cancel\frac{11 \: mm}{35 \: mm} = 0.3142$

$\small \boxed{ \sf \: R_f \: value \: of \: sample \: 1 =0.3142}$

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4 0

2 years ago