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3 years ago

7

life would not be possible on earth without the water that covers much of its surface, and the air that surrounds it. earrhs wat

er and air affect each other in many ways. which of the following situations represents an interaction between the atmosphere and the hydrosphere

1 answer:

maks197457 [2]3 years ago

4 0

Answer:evaporation from lakes and rivers

Explanation:I took the test

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What happens to liquid when it releases enough energy

Lelechka [254]

When a liquid releases enough energy<span>. the </span>liquid<span> freezes, changing to a solid.

Hope this answer helps! feel free to ask any additional questions :)</span>

8 0

4 years ago

Read 2 more answers

Which substance loses electrons in a chemical reaction

Ierofanga [76]

Answer:

Explanation:

Reducing agent

Reducing agent is an element or compound that loses or donates an electron to another chemical species in a redox chemical reaction. Since the reducing agent is losing electrons,it is said to have been oxidized.

8 0

3 years ago

What does ClF3 stand for

Nostrana [21]

Answer:

CHLORINE TRIFLUORIDE

Explanation:

3 0

3 years ago

Phosphorus pentachloride decomposes according to this equation. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) An equilibri

melisa1 [442]

Answer:

PCl₅ = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Cl₂ = 0.03X71 = 2.13 g

Explanation:

The equilibrium constant will remain the same irrespective of the amount of reactant taken.

Let us calculate the equilibrium constant of the reaction.

Kc= $\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}$

Let us calculate the moles of each present at equilibrium

$moles=\frac{mass}{molarmass}$

molar mass of PCl₅=208

molar mass of PCl₃=137

molar mass of Cl₂=71

moles of PCl₅ = $\frac{mass}{molarmass}=\frac{4.13}{208}=0.02$

moles of PCl₃= $\frac{mass}{molarmass}=\frac{8.87}{137}=0.06$

moles of Cl₂ = $\frac{mass}{molarmass}=\frac{2.90}{71}=0.04$

the volume is 5 L

So concentration will be moles per unit volume

Putting values

Kc = $\frac{\frac{0.06}{5}\frac{0.04}{5}}{\frac{0.02}{5}}=0.024$

Now if the same moles are being transferred in another beaker of volume 2L then there will change in the concentration of each as follow

$PCl_{5}--->PCl_{3}+Cl_{2}$

Initial 0.02 0.06 0.04

Change -x +x +x

Equilibrium 0.02-x 0.06+x 0.04+x

Conc. (0.02-x)/2 (0.06+x)/2 (0.04+x)/2

Putting values

0.024 = $\frac{(0.06+x)(0.04+x)}{(0.02-x)2}$

Solving

$(0.024(2)(0.02-x)=(0.06+x)(0.04+x)$

$0.00096-0.048x=0.0024+x^{2}+0.1x$

$0.148x+x^{2}+0.00144=0$

x = -0.01

so the new moles of

PCl₅ = 0.02 + 0.01 =0.03

PCl₃ = 0.06-0.01 = 0.05

Cl₂ = 0.04-0.01 = 0.03

mass of each will be:

mass= moles X molar mass

PCl₅ = 0.03 X 208 = 6.24g

PCl₃ = 0.05 X 137 =6.85 g

Cl₂ = 0.03X71 = 2.13 g

5 0

4 years ago

A common radio wavelength observed coming from astronomical objects is 21 cm. What temperature is associated with this radiation

qaws [65]

Answer:

The temperature associated with this radiation is 0.014K.

Explanation:

If we assume that the astronomical object behaves as a black body, the relation between its <em>wavelength</em> and <em>temperature</em> is given by Wien's displacement law.

$\lambda_{max}=\frac{b}{T}$

where,

λmax is the wavelength at the peak of emission

b is Wien's displacement constant (2.89×10⁻³ m⋅K)

T is the absolute temperature

For a wavelength of 21 cm,

$T=\frac{b}{\lambda _{max} } = \frac{2.89 \times 10^{-3} m.K }{0.21m} =0.014K$

8 0

4 years ago