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2 years ago

Please answer this questions sooni will mark him or her brillient

Chemistry

2 answers:

Misha Larkins [42]2 years ago

4 0

Answer:

1)Or

d.Sublimation

b.iodine is the solute and alcohol is solvent

a.compostion of solute

Snowcat [4.5K]2 years ago

3 0

Answer:

1) a) centrifugation

b) sublimation

2) a) iodine is the solute and alcohol is solvent

3) a) concentration of soLute

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the pressure and temperature of 10 liters of gas are doubled. if the original condition are 2 atmosphere of pressure and 400k wh

Harman [31]

Answer:

$\boxed{\text{10 L}}$

Explanation:

We have two pressures, two temperatures, and one volume.

This looks like a question in which we can use the Combined Gas Law to calculate the volume.

$\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}$

Data:

$\begin{array}{rclrclrcl}p_{1}& =& \text{2 atm}\qquad & V_{1} &= & \text{10 L}\qquad & T_{2}& =& \text{400 K}\\p_{2}& =& \text{4 atm}\qquad & V_{2} &= & \text{?}\qquad & T_{2}& =& \text{800 K}\\\end{array}$

Calculation:

$\begin{array}{rcl}\dfrac{2 \times 10}{400}& =& \dfrac{4V_{2} }{800}\\\\0.050& = &0.0050V_{2}\\V_{2}& = &\mathbf{10 L}\end{array}\\\text{The final volume is }\boxed{\textbf{10 L}}$

4 0

2 years ago

I need help with number 3. is the answer 1,2,3,or4

lubasha [3.4K]

Answer:

1 and 4

Explanation:

in both pictures the temperature is cooling down

8 0

3 years ago

The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C

Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})$

$1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})$

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})$

$1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})$

T₂=371.77 K= 98.62 °C

5 0

3 years ago

CORRECT ANSWER GETS BRAINLIST

pashok25 [27]

Answer:

B. halocline

Explanation:

it is a zone in the oceanic water that changes depending on the depth

Hope This Helped Sorry If Wrong

6 0

1 year ago

I need it please !!!

anyanavicka [17]

Answer:

im pretty sure its 2...

Explanation:

if its wrong im sorry

if its right brainliest pls?

6 0

2 years ago

Please answer this questions sooni will mark him or her brillient​

Please answer this questions sooni will mark him or her brillient