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Nastasia [14]
3 years ago
13

2.50 liter of a gas has a pressure of 165. kPa at 25.0°C. If the pressure increases to 600. kPa and the temperature to 100.0°C,

Chemistry
1 answer:
tiny-mole [99]3 years ago
7 0

Answer:

0.861 L

Explanation:

We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:

(P₁V₁)/T₁ = (P₂V₂)/T₂

Convert the temperatures to degrees Kelvin.

25.0°C -> 298 K, 100.0°C -> 373 K

Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:

(165(2.5))/298 = (600(V₂))/373

V₂ = (165(2.5)(373))/(298(600))

V₂ = 0.861 L

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GarryVolchara [31]

Answer:

K = 2.037*10^{-3} m/s

V_s = 0.0122 \ m/s

Explanation:

Given that;

diameter (d) = 10cm/2 = 0.1m/2 = 0.05 m

length (l) = 10 cm = 0.1 m

porosity = 50%

height (h) = 30 cm = 0.3 m

time (t) = 5 s

volume (v) = 60 cm³ = 60 × 10⁻⁶ m³

Q (flow rate) = \frac{v}{t}

Q = \frac{60*10^{-6} m^3}{5}

Q = 12*10^{-6} m^3 /sec

From constant head method, we use the relation K = (\frac{Q*L}{A*h}) to determine the hydraulic conductivity ; we have:

K = \frac{12*10^{-6}*0.1}{\frac{\pi}{4}0.05^2*0.3}

K = 0.002037\\\\K = 2.037*10^{-3} m/s

Seepage velocity V_s = \frac{velocity }{porosity}

where; velocity = K*i

=(2.037*10^{-3}*)(\frac{0.3}{0.1})

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V_s = \frac{6.111*10^{-3}}{0.5}

V_s = 0.0122 \ m/s

8 0
3 years ago
How many li are in 2.5 moles of li​
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Answer:

15.06 × 10²³ atoms of Li

Explanation:

Given data:

Number of moles of Li = 2.5 mol

Number of toms of Li = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 2.5 mol of Li:

1 mole of lithium = 6.022 × 10²³ atoms of Li

2.5 mol × 6.022 × 10²³ atoms of Li / 1 mol

15.06 × 10²³ atoms of Li

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Answer:

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