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2 years ago

What can form along faults when pushed together or pulled apart?

2 answers:

8 0

A Mountain is formed when two faults push together and a volcano is formed when two faults are pushed together and there is a pocket of air

SashulF [63]2 years ago

6 0

Answer:

Both

Explanation:

if you pull a mountain a part that would be great consequences and if a volcano breaks there will be fire .

You might be interested in

Metal or Non-metal?

goldenfox [79]

Answer:

Fluorine:non-metal

Bromine:non-metal

Hydrogen:non-metal

Beryllium:metal

Nitrogen:non-metal

5 0

3 years ago

If 850. mL of linseed oil has a mass of 620. g, calculate the density of linseed oil.

meriva

Answer:

<h3>The answer is 0.73 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 620 g

volume = 850 mL

We have

$density = \frac{620}{850} \\ = 0.7294117647...$

We have the final answer as

Hope this helps you

3 0

3 years ago

The ____ of the sun is the interior or center part of this star

Margaret [11]

Solar core I think... i hope it helps

5 0

3 years ago

Assume that temperature and number of moles of gas are constant in this problem.

IrinaVladis [17]

Question:

a. a direct linear relationship

b. an inverse linear relationship

c. a direct nonlinear relationship

d. an inverse nonlinear relationship

Answer:

The correct option is;

d. An inverse nonlinear relationship

Explanation:

From the universal gas equation, we have;

P·V = n·R·T

Where we have the temperature, T and the number of moles, n constant, therefore, we have

P×V = Constant, because, R, the universal gas constant is also constant, hence;

P×V = C

$P = \frac{C}{V}$

Since P varies with V then the graphical relationship will be an inverse nonlinear as we have

V P C

1 5 5

2 2.5 5

3 1.67 5

4 1.25 5

5 1 5

6 0.83 5

7 0.7 5

8 0.63 5

9 0.56 5

10 0.5 5

Where:

V = Volume

P = Pressure

C = Constant = 5

P = C/V

The graph is attached.

4 0

3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

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.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$