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Karo-lina-s [1.5K]
3 years ago
12

If A • B*2 = 1.8 x 10*-7, and C•B/D = 7.2 x 10*-4, find the value of A•D*2/C*2

Chemistry
1 answer:
joja [24]3 years ago
8 0

Answer:

A*D²/C² = 0.347

Explanation:

A*B² = 1.8*10⁻⁷----- (1)

making A subject of formula; A = 1.8 * 10⁻⁷/B² ----(2)

C* B/D = 7.2*10⁻⁴ ----(3)

making C subject of formula; C = D * 7.2*10⁻⁴/B ----(4)

substituting the values of A and C in the equation A*D²/C²

A*D²/C² = (D² * 1.8 * 10⁻⁷/B²)/(D * 7.2*10⁻⁴/B)²

A*D²/C² = (D² * 1.8 * 10⁻⁷/B²)/(D² * 5.184*10⁻⁷/B²)

A*D²/C² = D² * 1.8 * 10⁻⁷/B² * B²/D² * 5.184*10⁻⁷

A*D²/C² = 1.8 * 10⁻⁷/5.184*10⁻⁷

A*D²/C² = 0.347

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Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "
Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

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Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

<em>As Mass = density × volume</em>

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

<em>PV = nRt</em>

<em>V = nRT/P</em>

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0
3 years ago
If you have 12.5g of fluoride and 16.2g of sodium, which is the limiting reactant and how sodium fluoride in grams is your theor
Korvikt [17]

Answer:

F2 is the limiting reactant

27.6 grams of NaF is produced.

Explanation:

Balance the equation first.

2Na+ F2 ---> 2NaF

To find the limiting reactant, solve for how much NaF can be produced with Na and F2

12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)

=0.658 moles NaF

16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)

=0.705 moles NaF

Since F2 produced the least NaF, F2 is the limiting reactant.

Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.

0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF

27.6 moles of NaF would be theoretically produced.

8 0
3 years ago
Infrared (IR) and Nuclear Magnetic Resonance (NMR) are two spectroscopic techniques you've encountered in organic chemistry I, C
zalisa [80]

Answer:

The solution to this question can be defined as follows:

Explanation:

Please find the attached file for the solution:

8 0
3 years ago
The chief advantage of the metric system over other system of measurement is that it
kolbaska11 [484]

Answer:

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6 0
3 years ago
if a gas is held in 3.60 L, 298 K AND 800. mmHg, what is the new pressure of the gas if the volume us decreased by half and the
sleet_krkn [62]

Answer:

Explanation:

By Ideal Gas Law, P1*V1 / T1 = P2*V2 / T2

So new pressure = (P1*V1 / T1) / (V2 / T2)

= P1*V1*T2 / T1*V2

= 800*3.6*298 / 250*1.8

= 1907.2 mmHg

6 0
3 years ago
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