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4 years ago

Which mixture is heterogeneous?

1 answer:

6 0

A bowl of Fruit Loops cereal is a heterogeneous mixture because it has cereal bits of many colors floating around in milk.A bottle of balsamic vinaigrette salad dressing is a mixture that is heterogeneous, and has to be shaken up to make the mixture appear and taste more combined.Sand shaken up in a bottle of water is a heterogeneous mixture of sand particles floating around which will eventually settle to the bottom of the bottle, making it look a lot less like a mixture.

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I NEED THE ANSWER QUICKKK!!!

julsineya [31]

Answer: This is hard to do accurately, but here is my best assessment.

Explanation:

Experiment: B - describes how the experiment was done

Conclusion: A - The data support the prevailing hypothesis

Research: D - This is what we analyzed

Analysis: C - We compared the data

7 0

3 years ago

A town near the ocean would most likely have ________ humidity than a town near the mountains. (4 points) lower higher the same

aivan3 [116]

Higher humidity because there's more H2O in the air

3 0

3 years ago

Compounds X, C9H19Br, and Y, C9H19Cl, undergo base-promoted E2 elimination to give the same single alkene product, Z. Catalytic

Pavlova-9 [17]

Answer:

Compound X= 4-bromo-2,3,3-trimethylhexane

Compound Y= 5-chloro-2,3,3-trimethylhexane

Explanation:

The first step is set up the problem. That way we can obtain some clues. If we check figure 1 we can obtain some ideas:

-) If we have E2 reaction is not possible a methyl or hydride shift.

-) If we have an E2 reaction we will need an H in anti position to obtain the double bond. Therefore a double bond with the quaternary carbon (the carbon bonded to the 2 methyl groups).

The second step is to solve the alkene structure. We have to put the leaving group near to carbon that has more possible removable hydrogens. That's why the double bond is put it between carbons 5 and 4 of the alkane (Figure 2).

The third step is the structure of the alkyl bromide structure. To do this we have to check the alcohol produced by the alkene. In the hydration of alkanes reaction we will have a carbocation formation. Therefore we can have for the alkene proposed a methyl shift to obtain the most stable carbocation. With this in mind, we have to do the same for the Alkyl bromide that's why the Br is put it carbon 4 of the alkane. If we put the Br on this carbon we can have the chance of this methyl shift also, to obtain the same alcohol (figure 3).

Finally, for the alkyl chloride, we only have 2 choices because to produce the alkane we have to put the leaving group on one of the 2 carbons of the double bond. If we choose the same carbon on which we put the Br we can have the same behavior of the alkyl bromide (the methyl shift), therefore we have to put in the other carbon.

7 0

4 years ago

Hydrazine, , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine used as a fuel for rockets:

Tcecarenko [31]

Answer:

$1.25~mol~H_2O$ and $0.627~mol~N_2$

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ( $N_2H_4$ ) and oxygen ( $O_2$ ). So, we can start with the reaction between these compounds:

$N_2H_4~+~O_2~->~N_2~+~H_2O$

Now we can balance the reaction:

$N_2H_4~+~O_2~->~N_2~+~2H_2O$

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for $N_2H_4$ and 31.99 g/mol for $O_2$ ):

$20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4$

$20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2$

In the balanced reaction we have 1 mol for each reagent (the numbers in front of $O_2$ and $N_2H_4$ are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is $N_2H_4$ .

With this in mind, we can calculate the number of moles for each product. In the case of $N_2$ we have a 1:1 molar ratio (1 mol of $N_2$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2$

We can follow the same logic for the other compound. In the case of $H_2O$ we have a 1:2 molar ratio (2 mol of $H_2O$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O$

I hope it helps!

4 0

4 years ago

3. Describe how the water at the equator would be different from the water at the poles.

Dmitry_Shevchenko [17]

Answer:

there is a distinct decrease of salinity near the equator and at both poles, although for different reasons. Near the equator, the tropics receive the most rain on a consistent basis. As a result, the fresh water falling into the ocean helps decrease the salinity of the surface water in that region.

7 0

3 years ago