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3 years ago

At room temperature, one of the molecules is a gas and one is a liquid. Identify which structure is a gas and which is a liquid.

1 answer:

4 0

It starts at the bottom and then it goes up into the air as evaporation and that’s when the air gets cooler so gas is cooler and the liquid would be hotter

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A 250 mL sample of gas is collected over water at 35°C and at a total pressure of 735 mm Hg. If the vapor pressure of water at 3

ELEN [110]

Answer:

The volume of the gas sample at standard pressure is <u>819.5ml</u>

Explanation:

Solution Given:

let volume be V and temperature be T and pressure be P.

$V_1=250ml$

$V_2=?$

$P_{total}=735 mmhg$

1 torr= 1 mmhg

42.2 torr=42.2 mmhg

so,

$P_{water}=42.2mmhg$

$T_1=35°C=35+273=308 K$

Now

firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.

$P_{gas}=P_{total}-P_{water}$

=735-42.2=692.8 mmhg

Now

By using combined gas law equation:

$\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}$

$V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1$

$V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1$

Here $P_2 \:and\: T_2$ are standard pressure and temperature respectively.

we have

$P_2=750mmhg \:and\: T_2=273K$

Substituting value, we get

$V_2=\frac{692.8}{750}*\frac{273}{308} *250$

$V_2= 819.51 ml$

4 0

1 year ago

A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg

Brut [27]

Answer: $1.0\times 10^2g$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

$K_f$ = freezing point constant = $3.96^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

$7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}$

$x=1.0\times 10^2g$

Thus $1.0\times 10^2g$ urea was dissolved.

8 0

3 years ago

Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 10 ounces of water. Express y

d1i1m1o1n [39]

The pH of the solution in which one normal adult dose aspirin is dissolved is : 2.7

Given data :

mass of aspirin = 640 mg = 0.640 g

volume of water = 10 ounces = 0.295735 L

molar mass of aspirin = 180.16 g/mol

moles of aspirin = mass / molar mass = 0.00355 mol

<h3>Determine the pH of the solution </h3>

First step : <u>calculate the concentration of aspirin</u>

= moles of Aspirin / volume of water

= 0.00355 / 0.295735

= 0.012 M

Given that pKa of Aspirin = 3.5

pKa = -logKa

therefore ; Ka = $10^{-3.5}$ = $3.162 * 10^{-4}$

From the Ice table

$3.162 * 10^{-4}$ = $\frac{x + H^+}{[aspirin]}$ = $\frac{x^{2} }{0.012-x}$

given that the value of Ka is small we will ignore -x

x² = $3.162 * 10^{-4} * 0.012$

x = $1.948 * 10^{-3}$

Therefore

[ H⁺ ] = $1.948 * 10^{-3}$

given that

pH = - Log [ H⁺ ]

= - ( -3 + log 1.948 )

= 2.71 ≈ 2.7

Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is : 2.7

Learn more about Aspirin : brainly.com/question/2070753

4 0

1 year ago

What are three things that living things need to live

Grace [21]

Food shelter oxygen

8 0

3 years ago

Read 2 more answers

A sample of pure tin metal is dissolved in nitric acid to produce 15.00 mL of solution containing Sn2+. When this tin solution i

Rudik [331]

Answer:

1.00 M

Explanation:

Sn^2+ reacts with KMNO4 as follows;

5Sn^2+(aq) + 2MnO4^-(aq) + 16H^+(aq) ----> 5Sn^4+(aq) + 2Mn^+(aq) + 8H2O(l)

The number of moles of MnO4^- reacted = 42.1/1000 L × 0.145 mol/L

= 0.0061 moles

If 5 moles of Sn^2+ reacts with 2 moles of MnO4^-

x moles of Sn^2+ reacts with 0.0061 moles of MnO4^-

x= 5 × 0.0061/2

x= 0.015 moles

Since the volume of the Sn^2+ solution is 15.00mL or 0.015 L

number of moles = concentration × volume

Concentration = number of moles/volume

Concentration= 0.015 moles/0.015 L

Concentration = 1 M

6 0

3 years ago