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3 years ago

9

Water sits in an open beaker. assuming constant temperature and pressure, the vapor pressure of the water ______________ as the

water evaporates.

1 answer:

user100 [1]3 years ago

3 0

For a given system, vapor pressure can be defined as the pressure exerted by the vapor phase in equilibrium with the condensed phase.

In this case, as water sits in an open beaker it tends to evaporate thereby forming water vapor. This vapor phase then exerts a pressure over the liquid water, which is termed as the vapor pressure of water at that temperature and pressure. As water continues to evaporate, more and more molecules escape into the vapor phase which thereby increases the vapor pressure.

Ans: The vapor pressure of water increases as the water evaporates.

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Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

<u>Calculate the moles of potassium hydroxide.</u>

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

2 years ago

A 25.0 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution. The titration curve above was ob

Mnenie [13.5K]

Answer:

Weak acid

Explanation:

A titration curve is a graphical description of the change in pH of the solution in the conical flask as the reagent is added from the burette. A titration curve can be plotted for the different kinds of acid and base titrations. The volume of the titrant is always plotted as the independent variable and the pH of the solution as the dependent variable. The equivalence point is read off from the titration curve. A titration curve is very important because it shows the pH at various points during the titration.

A weak acid/strong base titration leads to an equivalence point above 7. From the question, we were told that the pH at equivalence point lies around 8. Hence the unknown substance must be a weak acid.

3 0

3 years ago

The reaction A + 2B occurs in one step in the gas phase. In each blank below, write the exponent of the concentration in the FOR

strojnjashka [21]

Answer:

see below

Explanation:

for A + 2B => Products ...

Rate Law => Rate =k[A][B]ˣ

As shown in expression, A & B are included, C is not.

6 0

2 years ago

Read 2 more answers

HELP PLS (CHEMISTRY)

IRINA_888 [86]

The first most obvious thing to note is when naming transitional metals, you have to state its charge with roman numerals (except for 1 if I remember correctly). For example, Iron (lll), iron has a charge of 3.

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2 years ago

Free-energy change, ΔG∘, is related to cell potential, E∘, by the equationΔG∘=−nFE∘where n is the number of moles of electrons t

mart [117]

Answer:

a) $\Delta G=372490 J$

b) $\Delta G=-568614 J$

Explanation:

a) The reaction:

$Mg(s) +Fe^{2+}(aq) \longrightarrow Mg^{2+}(aq) + Fe(s)$

The free-energy expression:

$\Delta G=-n*F*E$

$E=E_{red}-E_{ox]$

The element wich is reduced is the Fe and the one that oxidates is the Mg:

$-0.44V=E_{red}-(-2.37V)=1.93V$

The electrons transfered (n) in this reaction are 2, so:

$\Delta G=-2mol*96500 C/mol * 1.93 V$

$\Delta G=-372490 J$

b) If you have values of enthalpy and enthropy you can calculate the free-energy by:

$\Delta G=\Delta H - T* \Delta S$

with T in Kelvin

$\Delta G=-675 kJ*\frac{1000J}{kJ} - 298K*-357 J/K$

$\Delta G=-568614 J$

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2 years ago