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3 years ago

What is most likely to occur when jagged edges of rock plates grind past each other?

Chemistry

1 answer:

Ludmilka [50]3 years ago

7 0

Answer:

The most likely to occur when jagged edges of rock plates grind past each other is the presence of a high degree of frictional force.

This may cause the rocks to be broken down into smaller particles.

It also implies that the energy necessary for further disintegration and movement of rocks is stored up.

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a cylinder container is filled with air. its radius is 6cm and its height is 3.5cm. what is the volume of the air inside?

hram777 [196]

Answer:

Explanation:

I am math teacher

4 0

3 years ago

Vitamin K is involved in normal blood clotting. When 0.802 g of vitamin K is dissolved in 25.0 g of camphor, the freezing point

xenn [34]

Answer:

Molar mass of vitamin K = 450.56\frac{g}{mol}[/tex]

Explanation:

The freezing point of camphor = 178.4 ⁰C

the Kf of camphor = 37.7°C/m

where : m = molality

the relation between freezing point depression and molality is

Depression in freezing point = Kf X molality

Where

Kf = cryoscopic constant of camphor

molality = moles of solute dissolved per kg of solvent.

putting values

2.69°C = 37.7°C/m X molality

molality = 0.0714 mol /kg

$molality=\frac{molesofvitaminK}{massofcamphor(kg)}=\frac{moles}{0.025}$

moles of vitamin K = 0.0714X0.025 = 0.00178 mol

we know that moles are related to mass and molar mass of a substance as:

$moles=\frac{mass}{molarmass}$

For vitamin K the mass is given = 0.802 grams

therefore molar mass = $\frac{mass}{moles}=\frac{0.802}{0.00178}=450.56\frac{g}{mol}$

7 0

2 years ago

Question 11 please HELP

Lena [83]

The answer is A or B, I would put B.

6 0

3 years ago

A certain first-order reaction (A→products) has a rate constant of 8.10×10−3 s−1 at 45 ∘C. How many minutes does it take for the

stira [4]

Answer:

The answer is 5.7 minutes

Explanation:

A first-order reaction follow the law of $Ln [A] = -k.t + Ln [A]_{0}$ . Where [A] is the concentration of the reactant at any t time of the reaction, $[A]_{0}$ is the concentration of the reactant at the beginning of the reaction and k is the rate constant.

Dropping the concentration of the reactant to 6.25% means the concentration of A at the end of the reaction has to be $[A]=\frac{6.25}{100}.[A]_{0}$ . And the rate constant (k) is 8.10×10−3 s−1