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2 years ago

What evidence supports a scientist's conclusion that fossil B is older than fossil A? (4 points)

Chemistry

1 answer:

Hoochie [10]2 years ago

3 0

Answer:

Choice B, as time goes on the surface level rises and stuff buried underground gets further and further below surface level.

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Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H

solmaris [256]

Answer:

36.37% is the percent yield of the reaction.

Explanation:

$4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)$

1)0.650 L nitrogen gas , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T = 295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol$

2) Moles of ammonia gas= $\frac{2.53 g}{17 g/mol}=0.1488 mol$

Moles of oxygen gas = $\frac{3.53 g}{32 g/mol}=0.1101 mol$

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

$\frac{4}{3}\times 0.1101 mol=0.1468 mol$ of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

$\frac{2}{3}\times 0.1101 mol=0.0734 mol$ of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

$\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the reaction:

$\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%$

36.37% is the percent yield of the reaction.

3 0

3 years ago

does the nitro group on the pyridine ring make the ring more electron rich or more electron deficient

Arte-miy333 [17]

Answer:

more electron deficient

Explanation:

The nitro group is an electron withdrawing group. It withdraws electrons from the pyridine ring by resonance.

This electron withdrawal by resonance makes the pyridine ring less electron rich or more electron deficient.

Hence, the nitro group makes the pyrinde ring more electron deficient

3 0

3 years ago

Calculate the pH and pOH of 500.0 mL of a phosphate solution that is 0.285 M HPO42– and 0.285 M PO43–. (Ka for HPO42- = 4.2x10-1

jekas [21]

Answer: when concentrations of acid and base are same, pH = pKa

PH = 12.38 pOH = 1.62

Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00

8 0

3 years ago

A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What

NeTakaya

Answer:

Approximately $1.854\; \rm mol\cdot L^{-1}$ .

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

<h3>Formula mass of strontium hydroxide</h3>

Look up the relative atomic mass of $\rm Sr$ , $\rm O$ , and $\rm H$ on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

$\rm Sr$ : $87.62$ .
$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Calculate the formula mass of $\rm Sr(OH)_2$ :

$M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}$ .

<h3>Number of moles of strontium hydroxide in the solution</h3>

$M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}$ means that each mole of $\rm Sr(OH)_2$ formula units have a mass of $121.634\; \rm g$ .

The question states that there are $10.60\; \rm g$ of $\rm Sr(OH)_2$ in this solution.

How many moles of $\rm Sr(OH)_2$ formula units would that be?

$\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}$ .

<h3>Molarity of this strontium hydroxide solution</h3>

There are $8.71467\times 10^{-2}\; \rm mol$ of $\rm Sr(OH)_2$ formula units in this $47\; \rm mL$ solution. Convert the unit of volume to liter:

$V = 47\; \rm mL = 0.047\; \rm L$ .

The molarity of a solution measures its molar concentration. For this solution:

$\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}$ .

(Rounded to four significant figures.)

5 0

3 years ago

why do you think Kool Aid comes in packets that has a very finely ground powder (finely ground means very small particles).

marishachu [46]

Answer:

Answer 1:

When you pour the kool-aid into water, the little crystals go straight to the bottom because they are heavier than the water. If you left them there without stirring, and came back a few days later, you wouldn't see any crystals on the bottom. That's because the stuff in kool-aid can DISSOLVE in water, which means that each little molecule of kool-aid gets suspended between the molecules of water. When that happens, you can't see the kool-aid anymore...it's trapped between the water molecules. When you stir kool-aid, you help DISSOLVE the kool-aid in water by keeping all of the crystals off the bottom and in the water. So you see, stirring kool-aid speeds up the dissolving,

Answer 2:

Are you referring to Koolaid in the granular form?If so the koolaid grains sink in water because the grains have a greater density than that of water. Once your stir the grains dissolve and go into solution where they remain because the dissolved koolaid is miscible with water unlike oil (floats) or gasoline (sinks). How long did you let the koolaid remain in the water before you stirred it? I would think that if you left it undisturbed for a long time (days) it would eventually mix on its own.

Answer 3:

I'm not a chemist, but I think I can answer your question about Kool-Aid. Kool-Aid is mostly sugar, which is heavier than water, so when you pour it in it sinks to the bottom. When you stir it up the sugar (and flavoring) dissolves so that you don't have any solid particles any more. Stuff that is dissolved in water will not sink because it is no longer a physically separate thing. It becomes part of the water (or water-sugar-flavor solution). What happens if you pour the Kool-Aid in but don't stir it? Will it eventually dissolve? You may have to wait a long time, like over night. Try it and let me know what you find!

Answer 4:

It all has to do with the rate at which kool-aid crystals (basically its SUGAR!!) dissolves in water relative to the rate at which the sugar crystals sink. If you just dump the stuff in, it sinks because it is denser than the water. As it sinks it dissolves. But when you stir the water, the rate of dissolution becomes greater than the rate of sinking and so the crystals dissolve before they reach the bottom. So it all has to do with the comparison between the rate of sinking versus the rate of dissolution.

Now I have an experiment for you. What happens if you mix up some Jello and instead of letting it sit still, you keep stirring it??? WILL THE JELLO EVER SET??

You may have to borrow your mom's mixing machine because you will get tired of stirring after 10 minutes!!!!

If you do the experiment let me know how it turns out. Actually, you should set up a control. Make two batches of Jello...with one, put it in the refrigerator and dont stir; with the other, keep stirring it (in the refrigerator), if you can figure how to arrange that without your mom or dad getting mad!!!

8 0

3 years ago