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2 years ago

Which two processes turn layers of loose sediment into hard sedimentary rock? (ASAP)

Chemistry

1 answer:

kvasek [131]2 years ago

6 0

Answer:

compaction and cementation

Explanation:

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Exactly 10.0 L of air -25°C is heated to 100.0°C. What is the new volume if the pressure is kept constant?

zepelin [54]

Answer: V2= 15.0403226 Liters

Explanation:

Use V1/T1=V2/T2

Make sure you change the degrees Celsius to Kelvin. (Kelvin = degrees Celsius +273)

10.0L / 248 K = V2/ 373 K

Cross multiply V1 and T2 and divide by T1

(10.0 L)( 373K)/ 248 K = V2

V2= 15.0403226 Liters (Kelvin cancels out)

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3 years ago

I am the life cycle that contains tiny undeveloped plants,

LuckyWell [14K]

The water cycle regardless if it is in a lake, our bodies, food, or underground.

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2 years ago

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barxatty [35]

Answer: A

Explanation:

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2 years ago

How many moles of KBr will be produced from 10.51 moles of BaBr2?

gregori [183]

Answer:

21.02moles of KBr

Explanation:

Parameters given:

Number of moles BaBr₂ = 10.51moles

Complete reaction equation:

BaBr₂ + K₂SO₄ → KBr + BaSO₄

Upon inspecting the given equation, we find out that the atoms are not balanced on both sides of the equation:

The balanced equation is:

BaBr₂ + K₂SO₄ → 2KBr + BaSO₄

From the equation:

1 mole of BaBr₂ produces 2 moles of KBr

∴ 10.51 moles of BaBr₂ will yield (2 x 10.51) moles = 21.02moles of KBr

7 0

3 years ago

N2O5 decomposes to form NO2 and O2 with first-order kinetics. The initial concentration of N2O5 is 3.0 M and the reaction runs f

kow [346]

Answer : The final concentration of $N_2O_5$ is, 2.9 M

Explanation :

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $5.89\times 10^{-3}\text{ min}^{-1}$

t = time passed by the sample = 3.5 min

a = initial concentration of the reactant = 3.0 M

a - x = concentration left after decay process = ?

Now put all the given values in above equation, we get

$3.5=\frac{2.303}{5.89\times 10^{-3}}\log\frac{3.0}{a-x}$

$a-x=2.9M$

Thus, the final concentration of $N_2O_5$ is, 2.9 M

3 0

2 years ago