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xeze [42]
3 years ago
15

How are the processes of erosion and deposition closely related

Chemistry
1 answer:
denpristay [2]3 years ago
3 0

First of all, hey there! I hope you’re having a good day.

They are related because erosion moves broken materials and deposition means that the material is getting deposited, so that’s kind of how they are related. I hope you understand!!:)

Hope that helps :)

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What is the daughter nucleus of pt-199 after a beta decay?
Setler79 [48]

B. At-200

In beta decay the atomic number increases by 1

6 0
3 years ago
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
4 years ago
The density of lead is 11.4g/cm3. What volume, in ft3, would be occupied by 10.0 g of lead?
yawa3891 [41]

Answer:

3.10×10¯⁵ ft³.

Explanation:

The following data were obtained from the question:

Density (D) of lead = 11.4 g/cm³

Mass (m) of lead = 10 g

Volume (V) of lead =.?

Density (D) = mass (m) / Volume (V)

D = m/V

11.4 = 10 / V

Cross multiply

11.4 × V = 10

Divide both side by 11.4

V = 10 / 11.4

V = 0.877 cm³

Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:

1 cm³ = 3.531×10¯⁵ ft³

Therefore,

0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³

0.877 cm³ = 3.10×10¯⁵ ft³

Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.

Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.

7 0
3 years ago
Suppose that, from measurements in a microscope, you determine that a certain layer of graphene covers an area of 2.60μm2. conve
garri49 [273]
Given that 1 micrometer or micron (um) is equivalent by definition to 1 x 10^-6 m, this means that 1 square micron (um^2) is equivalent to (1 x 10^-6)^2 m^2, or 1 x 10^-12 m^2.

(2.60 um^2) * (1 x 10^-12 m^2 / 1 um^2) = 2.60 x 10^-12 m^2
Therefore the layer of graphene covers an area of 2.60 x 10^-12 m^2.
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5 0
3 years ago
Read 2 more answers
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grandymaker [24]
A. Because sodium is a metal and metals are good conductors
7 0
3 years ago
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