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3 years ago

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What is the mole fraction of Ba(OH)2 in an aqueous solution that contains 22.8% Ba(OH)2 by mass?

1 answer:

antoniya [11.8K]3 years ago

4 0

Answer:

0.03

Explanation:

22.8 g Ba(OH)2 (1 mol Ba (OH)2/ 171.34 g) = 0.133 mol Ba (OH)2

77.2 g H2O (1 mol H2O/18 g) = 4.29 mol H2O

X= molar fraction= mol Ba(OH)2/ mol total

X= 0.133/ (0.133+4.29) = 0.03

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At which temperature and pressure will a sample of neon gas behave most like an ideal gas?

Andreas93 [3]

Answer:

At STP, 760mmHg or 1 atm and OK or 273 degrees celcius

Explanation:

The standard temperature and pressure is the temperature and pressure at which we have the molecules of a gas behaving as an ideal gas. At this temperature and pressure, it is expected that the gas exhibits some properties that make it behave like an ideal gas.

This temperature and pressure conform some certain properties on a gas molecule which make us say it is behaving like an ideal gas. Ordinarily at other temperatures and pressures, these properties are not obtainable

Take for instance, one mole of a gas at stp occupies a volume of 22.4L. This particular volume is not obtainable at other temperatures and pressures but at this particular temperature and pressure. One mole of a gas will occupy this said volume no matter its molar mass and constituent elements. This is because at this temperature and pressure, the gas is expected to behave like an ideal gas and thus exhibit the characteristics which are expected of an ideal gas

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3 years ago

Read 2 more answers

Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.

Oduvanchick [21]

Answer:

103.9 g

Explanation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:

54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:

1.23 mol C₃H₈ * $\frac{3molCO_2}{1molC_3H_8}$ = 3.69 mol CO₂

We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:

3.69 mol CO₂ * 44 g/mol = 162.36 g

And <u>apply the given yield</u>:

162.36 g * 64.0/100 = 103.9 g

7 0

3 years ago

Calculate the dissociation constant of nh4oh(aq) if the degree of dissociation of 0.006 mol/kg solution is 0.053 and the activit

Anastasy [175]

The dissociation equation will be

NH4OH ---> NH4+ + OH-

Initial 0.006 0 0

Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053

Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053

Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053

Ka = 1.78 X 10^-5

7 0

3 years ago

Charlotte is driving at 58.6 mi/h and receives a text message. She looks down at her phone and takes her eyes off the road for 4

kramer

Answer: 19.71 feet

Explanation:

Given: Speed of Charlotte =58.6 mi/h

Since 1 hour = 3600 seconds

and 1 mile = 5280 feet

So, Speed of Charlotte = $\dfrac{58.6\times5280}{3600}\approx85.95$ mi/ sec

She looks down at her phone and takes her eyes off the road for 4.36 s.

Since , Distance = $\dfrac{speed}{time}$

So, Distance = $\dfrac{85.95}{4.36}=19.71\ feet$

Hence, Charlotte traveled 19.71 feet during this time.

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3 years ago

What is the limiting reactant for the following balanced equation when 9 moles of AlF3 are mixed with 12 miles of O2?

tamaranim1 [39]

<h2>Answer: $AlF_{3}$ </h2>

Explanation:

The chemical equation of the reaction that occurs when $AlF_{3}$ reacts with $O_{2}$ is

$4AlF_{3}+3O_{2}$ → $2Al_{2}O_{3}+6F_{2}$

$4$ moles of $AlF_{3}$ requires $3$ moles of $O_{2}$ .

$1$ mole of $AlF_{3}$ requires $\frac{3}{4}$ moles of $O_{2}$ .

Given that we have $9$ moles of $AlF_{3}$ .

$9$ moles of $AlF_{3}$ requires $\frac{3}{4}\times 9=6.75$ moles of $O_{2}$ .

But we have $12$ moles of $O_{2}$ .

So, $AlF_{3}$ will be consumed first.

So, $AlF_{3}$ is the limiting reagent.

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3 years ago