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2 years ago

5.1169 mol of Ne is held at 0.9148 atm and 911 K. What is the volume of its container in liters?

1 answer:

5 0

By applying the Boyle's equation and substituting our given data the volume of the container was found to be 418.14 Litres

<h3>Boyle's Law</h3>

Given Data

number of moles of Ne = 5.1169 mol

Pressure = 0.9148 atm

Temperature = 911 K

We know that the relationship between pressure and temperature is given as

PV = nRT

R = 0.08206

Making the volume subject of formula we have

V= nRT/P

Substituting our given data to find the volume we have

V = 5.1169*0.08206*911/0.9148

V = 382.522353554/0.9148

V = 418.14 L

Learn more about Boyle's law here:

brainly.com/question/469270

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If the caffeine concentration in a particular brand of soda is 1.87 mg/oz, drinking how many cans of soda would be lethal? Assum

Drupady [299]

2.77mg caffeine / 1oz12oz / 1canLethal dose: 10.0g caffeine = 10,000mg caffeine First, find how much caffeine is in one can of soda, then divide that amount by the lethal dose to find the number of cans. (2.77mg caffeine / 1oz) * (12oz / 1can) = 33.24mg caffeine / 1can. (10,000mg caffeine) * (1can / 33.24mg caffeine) = 300.84 cans. Since we can't buy parts of a can of soda, then we have to round up to 301 cans. Notice how all the values were set up as ratios and how the units cancelled.

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3 years ago

The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to

tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of $CuFeS_2$ = 183.511 g/mole

First we have to calculate the moles of Cu.

$\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles$

The moles of Cu = 4.7209 moles

From the given chemical formula, $CuFeS_2$ we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of $CuFeS_2$ = 4.4209 moles

Now we have to calculate the mass of $CuFeS_2$ .

Mass of $CuFeS_2$ = Moles of $CuFeS_2$ × Molar mass of $CuFeS_2$ = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of $CuFeS_2$ = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

3 0

3 years ago

How many grams are in 3.14 moles of PI₃?

OverLord2011 [107]

Answer:

$\boxed {\boxed {\sf 1290 \ g \ PI_3}}$

Explanation:

We want to convert from moles to grams, so we must use the molar mass.

<h3>1. Molar Mass</h3>

The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.

Phosphorus (P): 30.973762 g/mol
Iodine (I): 126.9045 g/mol

Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.

I₃: 126.9045 * 3=380.7135 g/mol

PI₃: 30.973762 + 380.7135 = 411.687262 g/mol

<h3>2. Convert Moles to Grams</h3>

Use the molar mass as a ratio.

$\frac {411.687262 \ g \ PI_3}{ 1 \ mol \ PI_3}$

We want to convert 3.14 moles to grams, so we multiply by that value.

$3.14 \ mol \ PI_3 *\frac {411.687262 \ g \ PI_3}{ 1 \ mol \ PI_3}$

The units of moles of PI₃ cancel.

$3.14 *\frac {411.687262 \ g \ PI_3}{ 1 }$

$1292.698 \ g\ PI_3$

<h3>3. Round</h3>

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.

1292.698

The 2 in the ones place tells us to leave the 9.

$1290 \ g \ PI_3$

3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>

4 0

3 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$