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Vladimir [108]
3 years ago
10

Water that flows back to the ocean is called

Chemistry
1 answer:
nordsb [41]3 years ago
4 0

Surface runoff

Explanation:

The water that flows back to the streams and oceans are called surface runoff.

Surface runoff is a component of the water cycle usually composed of water in the liquid form that flows back into oceans that are nearby.

  • The hydrologic cycle shows the cyclic process by which water passes in nature.
  • Water passes through different forms, solid, liquid and gases.
  • Surface runoff is water usually after rainfall that flows rapidly.
  • They move to the final basin of deposition usually joining up with other water sources.
  • This can be nearby streams, lakes or oceans.

learn more:

Downcutting a stream brainly.com/question/9259211

#learnwithBrainly

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he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
30. a. How do the properties of metals differ from those
Ahat [919]

Explanation:

Metals are elements that ionized by loss of electrons.

Ionic and molecular compounds are usually non-metals.

Properties of metals:

  • Metals have free mobile electrons and the metallic bonding ensures that.
  • They are usually electropositive and freely looses their electrons.
  • None of the metal is soluble without a chemical change occurring.
  • They are ductile and malleable.
  • Metals are good conductor or heat and electricity in their free uncombined state.
  • They are lustrous.

B. The specific property of metals accountable for their unusual electrical conductivity is due to the presence of free mobile electrons in their lattices.

learn more:

Metals brainly.com/question/2474874

#learnwithBrainly

3 0
3 years ago
What is specific heat capacity ​
LenaWriter [7]

Answer:

<u>A</u>

Explanation:

3 0
2 years ago
Read 2 more answers
A gold coin contains 3.47 × 10^23<br> gold atoms.<br> What is the mass of the coin in grams?
VashaNatasha [74]

Answer:

Mass = 114.26 g

Explanation:

Given data:

Number of gold atoms = 3.47×10²³ atoms

Mass in gram = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

3.47×10²³ atoms × 1 mol  /6.022 × 10²³ atoms

0.58 mol

Mass of gold:

Mass = number of moles × molar mass

Mass = 0.58 mol × 197 g/mol

Mass = 114.26 g

4 0
3 years ago
Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown
defon

The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

  • Find mol(mass : atomic mass)

Mn : 0.632 : 55 = 0.0115

O : 0.368 : 16 =0.023

  • Divide by the smallest mol(Mn=0.0115)

Mn : O =

\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2

The empirical formula : MnO₂

8 0
3 years ago
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