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3 years ago

Why are there fossils in sedimentary rocks?

Chemistry

1 answer:

melomori [17]3 years ago

8 0

Answer: Because, they form at temperatures and pressures that do not destroy fossil remains.

Explanation:

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‼️ i really need help. if anyone can do 10 pages of a grade 11 chemistry packet (multiple choice) let me know. i can pay money

Alja [10]

Answer:

mmmmmmmmmmmmmmmmmm

Explanation:

6 0

3 years ago

At 2000 ∘c the equilibrium constant for the reaction 2no(g)⇌n2(g)+o2(g) is kc=2.4×103. part a if the initial concentration of no

luda_lava [24]

Answer:

[NO] = 1.72 x 10⁻³ M.

Explanation:

For the reaction:

2NO(g) ⇌ N₂(g)+O₂(g),

Kc = [N₂][O₂] / [NO]².

At initial time: [NO] = 0.171 M, [N₂] = [O₂] = 0.0 M.

At equilibrium: [NO] = 0.171 M - 2x , [N₂] = [O₂] = x M.

∵ Kc = [N₂][O₂] / [NO]².

∴ 2400 = x² / (0.171 - 2x)² .

Taking the aquare root for both sides:

√(2400) = x / (0.171 - 2x)

48.99 = x / (0.171 - 2x)

48.99 (0.171 - 2x) = x

8.377 - 97.98 x = x

8.377 = 98.98 x.

∴ x = 8.464 x 10⁻².

∴ [NO] = 0.171 - 2(8.464 x 10⁻²) = 1.72 x 10⁻³ M.

∴ [N₂] = [O₂] = x = 8.464 x 10⁻² M.

3 0

3 years ago

A chemist titrates 110.0 mL of a 0.7684 M methylamine (CH3NH2) sotion with 0.4469 M HNO3 solution at 25 °C. Calculate the pH at

Firdavs [7]

Answer: The pH at equivalence point for the given solution is 5.59.

Explanation:

At the equivalence point,

$n_{HNO_{3}} = n_{CH_{3}NH_{2}}$

So, first we will calculate the moles of $CH_{3}NH_{2}$ as follows.

$n_{CH_{3}NH_{2}} = 0.764 M \times \frac{110 ml}{1000 ml/L}$

= 0.0845 mol

Now, volume of $HNO_{3}$ present will be calculated as follows.

Volume = $\frac{\text{no. of moles}}{\text{Molarity}}$

= $\frac{0.0845}{0.4469 M}$

= 0.1891 L

Therefore, the total volume will be the sum of the given volumes as follows.

110 ml + 189.1 ml

= 299.13 ml

or, = 0.2991 L

Now, $[CH_{3}NH_{3}^{+}] = \frac{0.0845 mol}{0.2991 L}$

= 0.283 M

Chemical equation for this reaction is as follows.

$CH_{3}NH_{3}^{+} + H_{2}O \rightleftharpoons CH_{3}NH_{2} + H_{3}O^{+}$

As, $k_{a} = \frac{k_{w}}{k_{b}}$

= $\frac{10^{-14}}{10^{-3.36}}$

= $2.29 \times 10^{-11}$

Now, $[HNO_{3}] = \sqrt{k_{a}[CH_{3}NH_{3}^{+}]}$

= $\sqrt{2.29 \times 10^{-11} \times 0.283}$

= $2.546 \times 10^{-6}$

Now, pH will be calculated as follows.

pH = $-log [H_{3}O^{+}]$

= $-log (2.546 \times 10^{-6})$

= 5.59

Thus, we can conclude that pH at equivalence point for the given solution is 5.59.

6 0

3 years ago

HELPPPPPPPPP 35 POINTSSSSSS!!!!!FILL IN THE BLANKS with either "reduce" or "oxidize"

masha68 [24]

reduce

oxidize

8 0

3 years ago

A sample of water in equilibrium with its vapor in a closed 3-L container has a vapor pressure of 23.8 torr at 25°C. The contain

QveST [7]

Answer:

23.8 torr

Explanation:

Vapor pressure of a liquid is simply the pressure of the vapor that arises when the liquid evaporates in a closed container above the liquid sample.

From the question given above, the vapor pressure is 23.8 torr at 25°C. Since the temperature is constant, the vapor will remain the same when equilibrium is reached in the new container as vapor pressure depends on the temperature of the liquid.

3 0

3 years ago

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