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Free_Kalibri [48]
3 years ago
6

During the processes of erosion and deposition, sediments that are the ________ in size will be carried the greatest distances b

efore being deposited
(There's no choices it's just fill in the blank)
Also
Most _______ rocks form under conditions found a few kilometers under earths surface
(No choices,fill in the blank)
Chemistry
2 answers:
Mila [183]3 years ago
4 0
<span>During the processes of erosion and deposition, sediments that are the smaller </span>in size will be carried the greatest distances before being deposited. Most metamorphic rocks form under conditions found a few kilometers under earths surface
Ahat [919]3 years ago
3 0

The full sentences are given below:

1. During the process of erosion and deposition, sediments that are the SMALLEST in size will be carried the greatest distance before being deposited.

Erosion and deposition are the methods by which sand and rock particles are moved from one place to another. The erosion can be caused by water or wind. Water and wind have the capacity to transport particle from one location and deposit them in another location. How far the erosion is able to move the particles depend on the weight of the particles. It is easier for erosion to carry small particles over a long distance than for it to carry large particles over the same distance.

2. Most METAMORPHIC rocks form under conditions found a few kilometer under the earth surface.

Metamorphic rocks generally are formed from existing rocks. The existing rocks are usually subjected to heat and pressure, which cause radical changes in the chemical and physical properties of the rock. Metamorphic rocks can be formed underneath the earth surface if they are subjected to high temperature and pressure by the rock layers above them.

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2. Which calorimeter lost energy?
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The kinetic energy bro
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3 years ago
What was the original pressure in a rigid container if the new pressure is 82.5atm and the temperature was raised from 44C to 67
zhenek [66]

Answer: 54 atm

Explanation:

I did 67/82.5 then got 0.8121212121. I them divided 44 by 0.81212121 and got 54.1791044776

4 0
3 years ago
450 nanometers to meters
lys-0071 [83]

Answer: 4.5 x 10e-7

Explanation: 450 x 1e+9 = correct answer

Multiply amount of nanometers by 1e+9 to get the approximate result in meters.

4 0
3 years ago
Reaction rate is expressed in terms of changes in the concentration of reactants and products. Write a balanced equation for the
KengaRu [80]

Answer : The balanced equations will be:

CH_4+2O_2\rightarrow 2H_2O+CO_2

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

Rate=-\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{d[CO_2]}{dt}

The balanced equations will be:

CH_4+2O_2\rightarrow 2H_2O+CO_2

5 0
3 years ago
In one experiment, the reaction of 1.00 mercury and an excess of sulfur yielded 1.16g of a sulfide of mercury
Nuetrik [128]

<u>Answer and Explanation:</u>

Mercury combines with sulfur as follows -

Hg + S = HgS

Hg = 200,59

S = 32,066 Therefore 1.58 g of Hg will react with -

1.58 multiply with 32,066 divide by 200,96 of sulfur.

= 0.25211 g S

This will form 1.58 + 0.25211 g HgS  = 1.83211 g HgS

The amount of S remaining = 1.10 - 0.25211  = 0.84789 g

5 0
3 years ago
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