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BigorU [14]
3 years ago
14

If 8.00 mol of an ideal gas at stp were confined to a cube, what would be the length in cm of an edge of this cube?

Chemistry
1 answer:
MatroZZZ [7]3 years ago
6 0
To solve this problem it is fundamentally, just look for the volume of the gas and convert it to cm3. At STP 1 mole = 22.4 liters. 8.00 moles x 22.4 liters/mole = 179.2 liters = 179,200 cm^3 Then. get the cube root of 179,200 cm^3. This would be equal to 56.38 cm and thus would be the length of the edge of this cube.
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Explanation:

The given data is as follows.

        V_{max} = 6.8 \times 10^{-10} \mu mol/min

          K_{m} = 5.2 \times 10^{-6} M

Now, according to Michaelis-Menten kinetics,

              V_{o} = V_{max} \times [\frac{S}{(S + Km)}]

where, S = substrate concentration = 10.4 \times 10^{-6} M

Now, putting the given values into the above formula as follows.

        V_{o} = V_{max} \times [\frac{S}{(S + Km)}]

        V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]

            V_{o} = 6.8 \times 10^{-10} \mu mol/min \times 0.667

                              = 4.5 \times 10^{-10} \mu mol/min

This means that V_{o} would approache V_{max}.

5 0
3 years ago
Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a
coldgirl [10]

The final temperature, t₂ = 30.9 °C

<h3>Further explanation</h3>

Given

24.0 kJ of heat = 24,000 J

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t₁= 25.5 °C

Required

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3 years ago
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3 years ago
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Given the balanced equation representing a reaction: 2na(s) + cl2(g) → 2nacl(s) + energy if 46 grams of na and 71 grams of cl2 r
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Answer is: 2) 117g.
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n(Na) = m(Na)÷M(Na) = 46g ÷ 23 g/mol = 2 mol.
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