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2 years ago

PLEASE HELP 50 POINTS!!!!!

Chemistry

1 answer:

OLEGan [10]2 years ago

8 0

Answer: This is what an effective conclusion should include.

Restate: Restate the lab experiment. Describe the assignment.

Explain: Explain the purpose of the lab. What were you trying to figure out or discover? ...

Results: Explain your results. ...

Uncertainties: Account for uncertainties and errors. ...

New: Discuss new questions or discoveries that emerged from the experiment.

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Using the periodic table, find the molecular mass of H2. H2 = g/mole

alekssr [168]

MH₂ = 2×mH = 2×1g = 2 g/mol

5 0

3 years ago

What is the mass of an object with a density of 3.4 g/mL and a volume of 500.0 mL

umka2103 [35]

Formula:
$mass=density *volume$

Given:
Density=3.4
Volume=500.0

Plug them into the formula:
$mass=3.4 *500.0=1700$

Final answer: 1700g

5 0

3 years ago

At 25 °C, how many dissociated OH– ions are there in 1243 mL of an aqueous solution whose pH is 2.07?

coldgirl [10]

<u>Answer:</u> The number of $OH^-$ ions dissociated are $8.57\times 10^{11}$

<u>Explanation:</u>

We are given:

pH = 2.07

Calculating the value of pOH by using equation, we get:

$2.07+pOH=14\\\\pOH=14-2.07=11.93$

To calculate hydroxide ion concentration, we use the equation to calculate pOH of the solution, which is:

$pOH=-\log[OH^-]$

We are given:

pOH = 11.93

Putting values in above equation, we get:

$11.93=-\log[OH^-]$

$[OH^-]=10^{-11.93}=1.17\times 10^{-12}M$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of solution = $1.17\times 10^{-12}M$

Volume of solution = 1243 mL = 1.243 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1.17\times 10^{-12}M=\frac{\text{Moles of }OH^-}{1.243L}\\\\\text{Moles of }OH^-=(1.17\times 10^{-12}mol/L\times 1.243L)=1.424\times 10^{-12}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, $1.424\times 10^{-12}mol$ number of $OH^-$ will contain = $(1.424\times 10^{-12}\times 6.022\times 10^{23})=8.57\times 10^{11}$ number of ions

Hence, the number of $OH^-$ ions dissociated are $8.57\times 10^{11}$

3 0

3 years ago

a) Whatis the composition in mole fractions of a solution of benzene and toluene that has a vapor pressure of 35 torr at 20 °C?

iogann1982 [59]

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.

For toluene and benzene would be:

$P_{B}=x_{B}*P_{B}^{o}$

$P_{T}=x_{T}*P_{T}^{o}$

Where:

$P_{B}$ is partial pressure for benzene in the liquid

$x_{B}$ is benzene molar fraction in the liquid

$P_{B}^{o}$ vapor pressure for pure benzene.

The total pressure in the solution is:

$P= P_{T}+ P_{B}$

And

$1=x_{B}+x_{T}$

Working on the equation for total pressure we have:

$P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}$

Since $x_{T}=1-x_{B}$

$P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}$

We know P and both vapor pressures so we can clear $x_{B}$ from the equation.

$x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}$

$x_{B}=\frac{35- 22}{75-22} = 0.24$

$x_{T}=1-0.24 = 0.76$

To get the mole fraction for the vapor we know that in the equilibrium:

$P_{B}=y_{B}*P$

$y_{T}=1-y_{B}$

$y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}$

$y_{B}=\frac{0.24*75}{35}=0.51$

$y_{T}=1-0.51=0.49$

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

7 0

3 years ago

How many moles of Al2O3 can be formed from 10.0 g of Al?

Nutka1998 [239]

Answer:

n Al= 10/27( mol)- >n Al2O 3 =5/27(mol)

Explanation:

3 0

2 years ago