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3 years ago

Why do not all rock types contain fossils?

Chemistry

1 answer:

balandron [24]3 years ago

4 0

Answer:

Igneous rocks form from molten rock, and rarely have fossils in them. Metamorphic rocks have been put under great pressure, heated, squashed or stretched, and fossils do not usually survive these extreme conditions. Generally it is only sedimentary rocks that contain fossils.

Explanation:

Hope it helps, some how.

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Use the solubility curve above to answer the following: If I add 130 g of potassium iodide to 100 g of water at 10°C, how would

zhenek [66]

<span>solution of KI becomes saturated at 10 degrees when around 135-138g KI are added to 100 g of water, so it should be still unsaturated, A. unsaturated (although it is very close to saturation)</span>

7 0

3 years ago

Complete the following radioactive decay problem.<br>222 86 RN to 4 2 HE + ?

Andreas93 [3]

Answer:

₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴

Explanation:

The given nuclear reaction shows alpha decay.

₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴

Properties of alpha radiations:

Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number less than 2 as compared to parent atom the starting atom.

Alpha radiations can travel in a short distance.

These radiations can not penetrate into the skin or clothes.

These radiations can be harmful for the human if these are inhaled.

These radiations can be stopped by a piece of paper.

₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + energy

8 0

3 years ago

A 2.20 mol sample of NO 2 ( g ) is added to a 3.50 L vessel and heated to 500 K. N 2 O 4 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) K c = 0.513

igor_vitrenko [27]

Answer:

[NO₂] = 0.434 M

[N₂O₄] = 0.0971 M

Explanation:

The equilibrum is: N₂O₄(g) ⇆ 2NO₂ (g)

1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.

Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:

2NO₂ (g) ⇆ N₂O₄(g)

Initially 2.20 mol -

React x x/2

X amount has reacted, and the half has been formed, according to stoichiometry.

Eq (2.20-x) / 3.50L (x/2)/ 3.50L

We divide by the volume because we need molar concentrations. Let's make the Kc's expression:

Kc = [N₂O₄] / [NO₂]²

0.513 = ((x/2)/ 3.50L) / [(2.20-x) / 3.50L]

0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)

0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50

0.203 - 0.184x + 0.0419x² = x/2 / 3.50

3.50(0.203 - 0.184x + 0.0419x²) = x/2

7 (0.203 - 0.184x + 0.0419x²) - x = 0

1.421 - 2.288x + 0.2933x² = 0 → Quadratic formula

a = 0.2933 ; b = -2.288 ; c = 1.421

(-b +- √(b²-4ac)) / (2a)

x₁ = 7.12

x₂ = 0.68 → We consider this value, so we can have a (+) concentration.

Concentrations in the equilibrium are:

[NO₂] = (2.20-0.68) / 3.50 = 0.434 M

[N₂O₄] = (0.68/2) / 3.50 = 0.0971 M

8 0

3 years ago

QUESTION 10

worty [1.4K]

I have no idea honestly I don’t remember I had it and I forgot it

6 0

3 years ago

To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water

Masteriza [31]

Answer:Sample Absorbance (625 nm)

A 0.536

B 0.783

C 0.045

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

4 0

3 years ago