Describe how to properly measure the volume of a liquid.

Ostrovityanka [42]

Answer:

Knowing this, researchers from the University of Southern Denmark decided to investigate the size of these hypothetical hidden particles. According to the team, dark matter could weigh more than 10 billion billion (10^9) times more than a proton.

Explanation:

If this is true, a single dark matter particle could weigh about 1 microgram, which is about one-third the mass of a human cell (a typical human cell weighs about 3.5 micrograms), and right under the threshold for a particle to become a black hole.

3 0

3 years ago

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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$

and,

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903$

Now we have to partial pressure of $C_2HBrClF_3$ and $O_2$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T$

$p_{C_2HBrClF_3}=0.0971\times 862torr=84torr$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.903\times 862torr=778torr$

Therefore, the partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

6 0

3 years ago

Predict the order of elution if you were to attempt to separate using column chromatography on silica gel and with a gradient el

Minchanka [31]

Answer:

Follows this order: B=> A => C.

Explanation:

NB: kindly check the attachment for the diagram of compounds A, B and C.

Elution is a very important concept in chromatography separation techniques. It deals with the use of eluent in the removal of an adsobate from an adsorbent. The principle behind Elution is just about how polar the solvent is.

So, in this question Compound B will go with the Elution first because of its polarity. Compound B has lesser polarity as compared to Compounds A and B.

Compound A will then elutes second because of its polarity too as resonance increases its polarity.

Last, compound C elutes because it has the highest polarity which is caused by electronegative atoms.

6 0

3 years ago

Note: Please show all work and calculation setups to get full credit. T. he following may be used on this assignment: specific h

olga55 [171]

Answer:

16974J of energy are required

Explanation:

The energy required is:

* The energy to heat solid water from -15°C to 0°C using:

q = m*S*ΔT

* The energy to convert the solid water to liquid water:

q = dH*m

* The energy required to increase the temperature of liquid water from 0°C to 75°C

q = m*S*ΔT

The first energy is:

q = m*S*ΔT

m = Mass water = 25g

S is specific heat of ice = 2.03J/g°C

ΔT is change in temperature = 0°C - (-15°C) = 15°C

q = 25g*2.03J/g°C*15°C

q = 761.3J

The second energy is:

q = dH*m

m = Mass water = 25g

dH is heat of fusion of water = 80cal/g

q = 80cal/g*25g

q = 2000cal * (4.184J/1cal) = 8368J

The third energy is:

q = m*S*ΔT

m = Mass water = 25g

S is specific heat of water= 4.184J/g°C

ΔT is change in temperature = 75°C-0°C = 75°C

q = 25g*4.184J/g°C*75°C

q = 7845J

The energy is: 7845J + 8368J + 761J =

16974J of energy are required

3 0

3 years ago

A reaction at −8.0°C evolves 854.mmol of boron trifluoride gas. Calculate the volume of boron trifluoride gas that is collected.

Bas_tet [7]

Answer: The volume of boron trifluoride gas that is collected is 18.6 L

Explanation:

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = 1 atm

V= Volume of the gas= ?

T= Temperature of the gas in kelvin = $-8.0^0C=273+(-8.0)=265K$

R= Gas constant = $0.0821Latm/Kmol$

n= moles of gas= 854 mmol = 0.854 mol (1mmol=0.001mol)

$1atm\times V=0.854\times 0.0821\times 265$

$V=18.6L$

Thus volume of boron trifluoride gas that is collected is 18.6 L

5 0

3 years ago