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netineya [11]
3 years ago
12

How will the vapor pressure of an aqueous solution of sodium chloride compare to that of pure water?

Chemistry
2 answers:
Eduardwww [97]3 years ago
5 0
The solutions vapor pressure would be lower.
slava [35]3 years ago
4 0

<u>Answer:</u> Vapor pressure of the solution will be less than the pure water.

<u>Explanation:</u>

Vapor pressure of a liquid is defined as the equilibrium pressure of the vapor above the liquid. It is also the pressure exerted by the evaporation of liquid.

When a solid is present in a liquid or a non-volatile substance is present in a liquid, the vapor pressure of the pure liquid tends to decrease.

The presence of a non-volatile solute in a solution reduces the escaping tendency of the solvent molecules into the vapor phase. This happens because solute particles occupy the position of solvent molecules on the liquid surface.

We are given an aqueous solution of sodium chloride and pure water.

Hence, vapor pressure of the solution will be less than the pure water.

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Need help with these two pls!!
denis23 [38]

what is a fire piston?

*a tool for starting a fire, likely invented in Ancient Southeast Asia.*

dont know the second one lol.

8 0
3 years ago
Calculate the mass of bromine in 50.0 g of potassium bromide
Elodia [21]

Answer:

6 Percent Composition. 1. Molar Mass of KBr K = 1(39.10) = 39.10 Br =1(79.90) =79.90 MM = 119.0 79.90 g        119.0 g = 0.6714 3. 0.6714 x 50.0g = 33.6 g Br 2.

4 0
2 years ago
The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat
Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

 Half life of lipase t_{1/2} = 8 min x 60 s/min

                                       = 480 s

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}&#10;

Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

Rate of hydrolysis V_m_{o} = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 5mmol/L

Therefore, time taken will be calculated as follows.

                    t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]

Now, putting the given values into the above formula as follows.

            t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]  

             = -\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s&#10;}{K_{M} ln (\frac{45 mmol/L&#10;}{9 mmol/L&#10;}) + (45 mmol/L - 9 mmol/L&#10;)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0
2 years ago
PLS HELP THE QUESTION IS ON THE PICTURE
IceJOKER [234]

<u>Concepts used:</u>

1 mole of an element or a compound has 6.022 * 10²³ formula units

So, we can say that: <em>Number of formula units = number of moles * 6.022*10²³</em>

number of moles of an element or a compound = given mass/molar mass

<u>__________________________________________________________</u>

<u>003 - </u><u>Number of CaH₂ formula units in 6.065 grams</u>

Number of Moles:

We know that the molar mass of CaH₂ is 42 grams/mol

Number of Moles of CaH₂ = given mass/molar mass

Number of moles = 6.065 / 42

Number of moles = 0.143 moles

Number of Formula units:

Number of formula units = number of moles * 6.022*10²³

= 0.143 * 6.022 * 10²³

= 0.86 * 10²³ formula units

__________________________________________________________

<u>004 </u><u>- Mass of 6.34 * 10²⁴ formula units of NaBF₄</u>

Number of Moles:

We mentioned this formula before:

<em>Number of formula units = number of moles * 6.022*10²³</em>

Solving it for number of moles, we get:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variable

Number of moles = 6.34 * 10²⁴ / 6.022*10²³

Number of moles=  10.5 moles

Mass of 10.5 moles of NaBF₄:

Molar mass of NaBF₄ = 38 grams/mol

Mass of 10.5 moles = 10.5 * molar mass

Mass of 10.5 moles = 10.5 * 38

Mass = 399 grams

__________________________________________________________

<u>005</u><u> - Number of moles in 9.78 * 10²¹ formula units of CeI₃</u>

Number of Moles:

We have the formula:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variables

Number of Moles = 9.78 * 10²¹ / 6.022*10²³

Number of Moles = 1.6 / 10²

Number of Moles = 1.6 * 10⁻² moles   OR   0.016 moles

3 0
2 years ago
Ammonia (nh3) is widely used as a fertilizer and in many household cleaners. how much ammonia is produced when 6.64 mol of hydro
olga55 [171]
N₂ + 3H₂ ⇒ 2NH₃

doesnt matterN₂ + 6.64H₂ ⇒ 2NH₃

(6.64H₂/3H₂) x (2NH₃) =4.4266667

rounded to sig figs= 4.43

5 0
2 years ago
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