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1 year ago

What would happen if evaporation does not take place in nature

Chemistry

1 answer:

ivann1987 [24]1 year ago

6 0

Answer:

Evaporation is an important process in the water cycle. If there is no evaporation, there is no moisture in the atmosphere. As a result, clouds cannot form and there is no rain. Plants do not receive natural water and die. And there would be no rivers or lakes because their water comes from evaporation via rain or snow.

Explanation:

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A strontium-90 atom that has lost 2 electrons has ________ protons, ________ neutrons, and ________ electrons.

Sliva [168]

An atom of strontium-90 contains how many electrons, protons, and neutrons?
38e⁻, 38 protons, and 52 neutrons

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3 years ago

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An exacuted glass vessel weighs 50 g when empty, 148g when filled with an liquid of density o.989/cc and 50.5 g whenfilled with

Zanzabum

MW of gas : 124.12 g/mol

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of density

The unit of density can be expressed in g/cm³ or kg/m³

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

glass vessel wieight = 50 g

glass vessel + liquid = 148 ⇒ liquid = 148 - 50 =98 g

volume of glass vessel :

$\tt V=\dfrac{m}{\rho}=\dfrac{98}{0.989}=99.1~ml$

An ideal gas :

m = 50.5 - 50 = 0.5 g

P = 760 mmHg = 1 atm

T = 300 K

$\tt PV=\dfrac{mass(m)}{MW}.RT\\\\MW=\dfrac{m.RT}{PV}\\\\Mw=\dfrac{0.5\times 0.082\times 300}{1\times 0.0991}=124.12~g/mol$

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3 years ago

The coolest stars in the sky are ____ in color

Mekhanik [1.2K]

Red is the answer in the blank

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3 years ago

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Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution i

alexdok [17]

Answer:

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$ .

Explanation:

The pH of the solution = 2.46

$pH=-\log[H^+]$

$2.46=-\log[H^+]$

$[H^+]=0.003467 M$

$HA\rightleftharpoons H^++A^-$

Initially

0.0144 0 0

At equilibrium

(0.0144-x) x x

The expression if an dissociation constant is given by :

$K_a=\frac{[A^-][H^+]}{[HA]}$

$K_a=\frac{x\times x}{(0.0144-x)}$

$x=[H^+]=0.003467 M$

$K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}$

$K_a=1.099\times 10^{-3}$

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$ .

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3 years ago

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

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3 years ago

What would happen if evaporation does not take place in nature​

What would happen if evaporation does not take place in nature