1pt Which form of energy can be found in an AA battery?

What are the correct steps for evolution by natural selection?

katrin [286]

(っ^-^ )っHey There!

Answer → The five steps of the process of natural selection are variation, inheritance, selection, time and adaptation

✨Hope this Helps✨

8 0

3 years ago

The reaction between iron(II) oxide and carbon monoxide produces iron and carbon dioxide. How many moles of iron can be obtained

My name is Ann [436]

Answer:

4.75 moles of Fe

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

FeO + CO —> Fe + CO2

Now, we can determine the number of mole of iron, Fe produced by the reaction of 4.75 mol of FeO with excess CO as follow:

From the balanced equation above,

1 mole of FeO reacted to produce 1 mole of Fe.

Therefore, 4.75 moles of FeO will also react to produce 4.75 moles of Fe.

Therefore, 4.75 moles of Fe is produced.

4 0

3 years ago

. What is a carbonyl group? Draw the carbonyl groups that are characteristic of aldehydes, ketones, carboxylic acids, and esters

Elan Coil [88]

Answer :

Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as $C=O$

Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.

The general formula of carboxylic is, $C_nH_{2n+1}COOH$ . According to the IUPAC naming, the carboxylic are named as alkanoic acids.

Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.

The general representation of aldehyde is, $R-CHO$ . According to the IUPAC naming, the aldehyde are named as alkanals.

Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.

The general representation of ketone is, $R-CO-R^'$ . According to the IUPAC naming, the ketone are named as alkanone.

Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.

The general representation of ester is, $R-COO-R^'$ . According to the IUPAC naming, the ester are named as alkyl alkanoate.

6 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

4 years ago

This diagram shows the forces acting on a car. The center dot represents the car, and the arrows represent the forces acting on

Mariulka [41]

Answer : The correct option is B- Towards the left.

Solution : Given,

Force on car in upward direction, $F_1$ = 6000 N

Force on car in downward direction, $F_2$ = 6000 N

Force on car in right direction, $F_3$ = 5000 N

Force on car in left direction, $F_4$ = 6000 N

Force : Force is equal to the mass multiplied by the acceleration. The force is directly proportional to the acceleration.

Formula : F = m × a

where, F is force, m is mass and a is acceleration.

From the given diagram we conclude that the $F_1$ and $F_2$ are in opposite direction and the force acting on car are same. So, their net force is zero.

And in case of $F_3$ and $F_4$ , the net force will not be zero because force acting on car are different. The car will accelerate towards left direction because $F_4$ is higher than the $F_3$ .

So, the car will accelerate towards left direction.

3 0

3 years ago