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jasenka [17]
3 years ago
13

The rock outcrop, shown above demonstrates the law of included fragments. Identify the correct definition for the law of include

d fragments.
Any pieces of rock found in another layer are OLDER than the layer in which they are found.

Any pieces of rock found in another layer are YOUNGER than the layer in which they are found.

Any pieces of rock found in another layer are THE SAME AGE as the layer in which they are found.

Any pieces of rock found in another layer are not able to be relative age dated.
100 points if you answer correctly

Chemistry
1 answer:
Radda [10]3 years ago
7 0

Answer:

A

Explanation:

The law of included fragments states: Fragments found in a rock must be older than the rock itself.

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Does chocolate milk come from brown cows?
ss7ja [257]
Chocolate milk is just regular white milk that has chocolate syrup/powder mixed into it
So no, chocolate milk does not come from brown cows

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11 Which compound has the greatest percent composition by mass of sulfur?(1) BaS (3) MgS
harina [27]
Percent composition by mass is calculated (mass of element within compound)/(mass of compound)*100. The lower the total molar mass of the compound, the greater the percent composition of sulfur. In this case, MgS would be that compound, since Mg has the lowest molar mass of the four elements bonded to S.
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A formic acid buffer solution contains 0. 20 m h c o o h hcooh and 0. 24 m h c o o − hcoox−. the pka of formic acid is 3. 75. wh
irina1246 [14]

A buffer solution contains an equivalent amount of acid and base. The pH of the solution with an acid dissociation constant (pKa) value of 3.75 is 3.82.

<h3>What is pH?</h3>

The amount of hydrogen or the proton ion in the solution is expressed by the pH. It is given by the sum of pKa and the log of the concentration of acid and bases.

Given,

Concentration of salt [HCOO⁻] = 0.24 M

Concentration of acid [HCOOH] = 0.20 M

The acid dissociation constant (pKa) = 3.75

pH is calculated from the Hendersons equation as,

pH = pKa + log [salt] ÷ [acid]

pH = 3.75 + log [0.24] ÷ [0.20]

= 3.75 + log (1.2)

= 3.75 + 0.079

= 3.82

Therefore, 3.82 is the pH of the buffer.

Learn more about pH here:

brainly.com/question/27181245

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6 0
2 years ago
What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
Name the separation technique which could be used to separate mud and water without heating​
Westkost [7]

Answer:

Decantation

Explanation:

Decantation is a process to separate

mixtures by removing a liquid layer that

is free of a precipitate, or the solids

deposited from a solution. The purpose

may be to obtain a decant (liquid free

from particulates) or to recover the

precipitate.

Decantation relies on gravity to pull

precipitate out of the solution, so there is

always some loss of product, either from

the precipitate not fully falling out of the

solution or from liquid remaining when

separating it from the solid portion.

The Decanter

A piece of glassware called a decanter is

used to perform decantation. There are

several decanter designs. A simple

version is a wine decanter, which has a

wide body and a narrow neck. When

wine is poured, solids stay in the base of

the decanter.

In the case of wine, the solid is usually

potassium bitartrate crystals. For

chemistry separations, a decanter may

have a stopcock or valve to drain the

precipitate or dense liquid, or it may

have a partition to separate fractions.

Use of Alum and filtration process can also be administered for further purification

8 0
3 years ago
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