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Rashid [163]
3 years ago
10

PLEASE HELP!CHEMISTRY QUESTION

Chemistry
1 answer:
ozzi3 years ago
8 0

<u>Answer:</u> The partial pressure of nitrogen on Venus is 81 mmHg

<u>Explanation:</u>

To calculate the partial pressure of the gas, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}

where,  

p_A = partial pressure of nitrogen gas = ?

p_T = total pressure = 2700 mmHg

\chi_A = mole fraction of nitrogen gas = 3.0 % = 0.03

Putting values in above equation, we get:

p_{N_2}=2700mmHg\times 0.03\\\\p_{N_2}=81mmHg

Hence, the partial pressure of nitrogen on Venus is 81 mmHg

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5 0
3 years ago
What is the answer for number one please help
patriot [66]

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4 0
4 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

3 0
3 years ago
What is the minimum frequency of light required to observe the photoelectric effect on pd
Tems11 [23]

Answer:

1.26 × 10¹⁵ s⁻¹

Explanation:

Work function is the minimum energy required to remove an electron from the surface of metal

energy of the electron = hf - Φ

Φ = work function = hf₀ where f₀  = threshold frequency

f₀ = Φ / h  where h ( Planck constant = 6.626 × 10⁻³⁴ Js)

Φ = 5.22eV = 5.22 × 1 eV  where 1 eV = 1.60217662 × 10⁻¹⁹ J

Φ = 5.22 × 1.60217662 × 10⁻19 J = 8.363362 × 10⁻¹⁹ J

f₀ =  (8.363362 ×10⁻¹⁹  J) / (6.626× 10⁻³⁴ Js) = 1.26 × 10¹⁵ s⁻¹

The frequency must be greater than the 1.26 × 10¹⁵ s⁻¹ to observe the emission

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4 years ago
Was is meant by the term suprenatural? Does science deal with the supernatural?
marshall27 [118]

Answer:

it's option D

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