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vovikov84 [41]
3 years ago
8

Question 4 (1 point)

Chemistry
2 answers:
Alex3 years ago
8 0

Answer:

Wusup,

Your answer is B: Larger than the moon

hope that helps, tell me if u get it right.

Explanation:

Brilliant_brown [7]3 years ago
8 0

Answer:

I put B and it was incorrect, sadly I dont know the correct answer

Explanation:

You might be interested in
How many different bases of DNA​
const2013 [10]

Answer:

four

Explanation:

adenine, thymine, guanine, and cytosine

4 0
3 years ago
Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3NO2(g)+H2O(l)→2HNO3(l)+
Vera_Pavlovna [14]

Answer:

5 moles of NO₂ will remain after the reaction is complete

Explanation:

We state the reaction:

3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g)

3 moles of nitric oxide can react with 1 mol of water. Ratio is 3:1, so we make this rule of three:

If 3 moles of nitric oxide need 1 mol of water to react

Then, 26 moles of NO₂ may need (26 .1) / 3 = 8.67 moles of H₂O

We have 7 moles of water but we need 8.67 moles, so water is the limiting reactant because we do not have enough. In conclusion, the oxide is the reagent in excess. We can verify:

1 mol of water needs 3 moles of oxide to react

Therefore, 7 moles of water  will need (7 .3)/1 = 21 moles of oxide

We have 26 moles of NO₂ and we need 21, so we still have oxide after the reaction is complete. We will have (26-21) = 5 moles of oxide that remains

5 0
2 years ago
A hiker climbs a hill. How does the potential energy of the hiker change as she climbs higher?
zloy xaker [14]
The answer should be A
6 0
3 years ago
A piece of metal with a mass of 611 g is placed into a graduated cylinder that contains 25.1 ml of water, raising the water leve
jeka94

Mass of metal piece is 611 g and volume of graduated cylinder is 25.1 mL. When metal piece is placed in the graduated cylinder water level increases to 56.7 mL. The increase in volume is due to volume of metal piece that gets added to the volume of water.

Thus, volume of metal piece can be calculated by subtracting initial volume from the final one.

V_{metal}=V_{final}-V_{initial}=(56.7-25.1)mL=31.6 mL

Thus, volume of metal piece will be 31.6 mL. The mass of metal piece is given 611 g, density of metal can be calculated as follows:

d=\frac{m}{V}=\frac{611 g}{31.6 mL}=19.33 g/mL

Therefore, density of metal is 19.33 g/mL.

3 0
3 years ago
stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,
Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

8 0
3 years ago
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