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uranmaximum [27]
3 years ago
7

The mole fraction of nitrogen in the air is 0.7808. this means that 78.08% of the molecules in the air are nitrogen. when the at

mospheric pressure is 760 torr, the partial pressure of nitrogen in the air is _______.
Chemistry
1 answer:
Pachacha [2.7K]3 years ago
7 0

Answer:

p = \boxed{\text{593 torr}}

Explanation:

For this question, we must use Dalton's Law of Partial Pressures:

The partial pressure of a gas in a mixture of gases equals its mole fraction times the total pressure:

p = \chi p_{\text{tot}}

Data:

χ = 0.7808

p_{\text{tot}} = \text{ 760 torr}

Calculation:

p = 0.7808 \times \text{ 760 torr}\\\\p= \boxed{\textbf{593 torr}}

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iragen [17]

Answer:

experiment can show the movement of pollutants through the groundwater!

Explanation:

8 0
3 years ago
My swimming pool is rectangular (16 feet by 34 feet) and has a depth of 6 feet. Lets imagine that my pool water is full to the t
Reil [10]

Answer:

Number of moles of photons required = 5.04 × 10⁴ moles

Explanation:

The energy of a photon can be calculated from Planck's equation E = hc/λ

Where h = 6.63 × 10-³⁴ Js and c, the velocity of light = 3.0 × 10⁸ m/s

Energy of one mole of photons = N₀ × hc/λ

wavelength of photon, λ = 520 nm = 5.20 × 10-⁷ m

Energy of one mole of photons = 6.02 × 10²³ × 6.63 × 10−³⁴ × 3 × 10⁸/5.20 × 10-⁷

Energy of one mole of photons = 2.30 × 10⁵ J/mol

Energy required to raise the temperature of a given mass of a substance, E = mcΔT

Where m is mass of substance,  c is specific heat capacity,  ΔT is temperature difference

Mass ofnwternin the pool = volume × density

Volume of water = Volume of swimming pool

Volume of water = 16 × 34 × 6 ft³ = 3264 ft³

1 ft³ = 28316.8 cm³; 3264 ft³ = 28316.8 × 3264 = 92426035.2 cm³

Density of water = 1 g/cm³

Mass of water = 92426035.2 cm³ × 1 g/cm³ = 92426035.2g

ΔT = 80°C - 50°C = 30°C, c = 4.18 J/g/K

Energy required to raise 92426035.2 g water by 30° C = 92426035.2 × 4.18 × 30

Energy required = 1.16 × 10¹⁰ J

Hence, number of moles of photons required = 1.16 × 10¹⁰ J/2.30 × 10⁵ J/mol

Number of moles of photons required = 5.04 × 10⁴ moles

5 0
3 years ago
C12H7Cl3FNaO2 what are them elements in that chemical formula
LuckyWell [14K]
Assuming you are asking for the names of the elements in that formula , the answer is

carbon
hydrogen
chlorine
fluorine
sodium
oxygen
8 0
4 years ago
Can someone help me with this?
amm1812

Answer:

dnaq tester

Explanation:

5 0
3 years ago
If the half-life of Carbon-14 is 5700 years, how many years would it take a sample to decay from 1 gram to 31.3 mg
andrew-mc [135]

Answer:

28500 years

Explanation:

Applying,

A = A'(2^{x/y})............... Equation 1

Where A = Original mass of Carbon-14, A' = Final mass of carbon-14 after decaying, x = total time, y = half-life.

From the question,

Given: A = 1 g, A' = 31.3 mg = 0.0313 g, y = 5700 years.

Substitute these values into equation 1

1 = 0.0313(2^{x/5700})

2^{x/5700} = 1/0.0313

2^{x/5700}  = 31.95

2^{x/5700} ≈ 32

2^{x/5700} ≈ 2⁵

Equating the base and solve for x

x/5700 ≈ 5

x ≈ 5×5700

x ≈ 28500 years

3 0
3 years ago
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