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defon
3 years ago
14

A chemist is identifying the elements present in a sample of seawater. What characteristic of an element's atoms always determin

es the element's identity?
Group of answer choices (Could someone help me? I am kind of confused) Thank you so much

A. The number of valence electrons

B. The location of valence electrons

C. The number of neutrons

D. The number of protons
Chemistry
1 answer:
eduard3 years ago
3 0
B explanation explanation point
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Balance the following reaction _ Al + _ Cl, → _ AICI,
Free_Kalibri [48]

Answer:

1 Al + 1 Cl -> 1 AlCl

Explanation:

It is already balanced.

3 0
3 years ago
Which rules are violated in the following electronic configuration 1s2 2s3​
aalyn [17]
There is no element in 2s3
5 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
4 years ago
0:00 一<br> what is 18,000 divided by 450,000 in scientific notation
ankoles [38]

Answer:

4 × 10-2

(scientific notation)

= 4e-2

(scientific e notation)

= 40 × 10-3

(engineering notation)

(thousandth; prefix milli- (m))

Explanation:

All of the anwsers

3 0
3 years ago
Exercise A mixture of 250 mL of methane, CH4, at 35 °C and 0.55 atm and 750 mL of propane, C3Hg, at 35° C and 1.5 atm, were intr
dalvyx [7]

Explanation:

The given data is as follows.

      V_{1} = 250 mL,     V_{2} = 750 mL

      T_{1} = 35^{o}C = 35 + 273 K = 308 K

      T_{2} = 35 + 273 K = 308 K

      P_{1} = 0.55 atm,    P_{2} = 1.5 atm

               P = ? ,         V = 10.0 L

Since, temperature is constant.

So,    P_{1}V_{1} + P_{2}V_{2} = PV

Now, putting the given values into the above formula as follows.

         P_{1}V_{1} + P_{2}V_{2} = PV

         0.55 atm \times 250 mL + 1.5 atm \times 1.5 atm = P \times 10.0 L

                     P = 0.126 atm

As, 1 atm = 760 torr. So, 0.126 atm \times \frac{760 torr}{1 atm} = 95.76 torr.

Thus, we can conclude that the final pressure, in torr, of the mixture is 95.76 torr.

8 0
3 years ago
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