Organic Chemistry

What type of bond does sodium and bromine form?

vitfil [10]

Sodium is a metal and bromine is a nonmetal so they form an ionic compound
nonmetals and nonmetals form covalent compounds

7 0

3 years ago

How do CFC’s illustrate that technical advances can be a “double-edged” sword?

hram777 [196]

Answer:

CFC’s illustrate that technical advances can be a “double-edged” sword as CFC's were made for the advantages but soon or later it came as a disadvantage for environment.

CFC's was discovered with many advantages such as they are stable, non-corrosive, and replacement of harmful ammonia used in refrigerators and air conditioning systems. But soon it is also discovered that CFC's are destroying the upper atmosphere layer called ozone, which is a very big disadvantage for us. CFCs are major contributor to global warming after carbon dioxide.

Hence, it proves that CFC’s shows how technological advancement can be “double-edged” sword.

7 0

3 years ago

Complete the following reaction. 14N+ on →?C+1H 18C 14 1 C 18C

CaHeK987 [17]

Answer: The complete reaction is as follows.

$^{14}_{7}N + ^{1}_{0}n \rightarrow ^{14}_{6}C + ^{1}_{1}H$

Explanation:

When nucleus of two or more atoms are bombarded together then it leads to the formation of new particles with new identity. This type of reaction are called nuclear reaction.

For example, $^{14}_{7}N + ^{1}_{0}n \rightarrow ?C + ^{1}_{1}H$

Here, nitrogen atom when bombarded with a neutron then it is forming hydrogen and a carbon atom.

As total atomic mass on reactant side is (14 + 1) = 15

So, the atomic mass of carbon formed on product side is (15 - 1) = 14.

The number of protons holded by this carbon atom is (7 - 1) = 6.

Therefore, we can conclude that the complete reaction is as follows.

$^{14}_{7}N + ^{1}_{0}n \rightarrow ^{14}_{6}C + ^{1}_{1}H$

6 0

3 years ago

How can I balance this equation? TELL ME HOW you did it and all the steps and you will be the brainliest.

lianna [129]

Answer:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Explanation:

To balance, start of with the groups that are common on both sides of the reaction equation;

In this case, these are the SO4 groups treating them as single units;

There are 3 on the right side and 1 on the left side, so we put 3 in front of the H2SO4 to balance this first;

Next, deal with the Cr, there are 2 Cr on the right side and 1 on the other side, so we put a 2 in front of the H2CrO4 to balance that;

Thirdly, we notice the C are already balanced as there are 2 on each side so this is fine;

Lastly, we can deal with the O and H;

Bearing in mind the numbers that are in front of the molecules now from prior balancing, there are 9 O and 16 H on the left side and 3 O and 5 H on the right;

7 H2O on the right side would balance the O, but gives us 18 H, which is 2 too many H;

If we were to put 2 in front of the two organic molecules (the ones with C) on either side, we would balance the O by having 6 H2O, but this gives 2 fewer H than necessary;

In order for the H to balance, we need to have 13/2 (or 6.5) H2O, which means we need 3/2 (or 1.5) in front of each organic molecule;

Since, it is not sensible to have 13/2 water molecules or 3/2 organic molecules, we can just multiply everything by 2;

Thus we end up with:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Rules of thumb:

- When there are common chemical groups (e.g. SO4) on both sides of a reaction equation, treat them as single units

- Start of with balancing these common groups

- Thereafter, balance the atoms that appear in only one reactant and one product

- Proceed to balance the atoms that appear in more than one reactant or product

- Typically, you should deal with the O and H last

4 0

3 years ago

How many grams of ag can be formed from 5.50 grams of ag2o in the equation: 2ag2o (s) → 4ag (s) o2 (g)?

oee [108]

From the given balanced equation we have find out the amount (in gm) of Ag formed from 5.50 gm of Ag₂O.

2Ag₂O(s) → 4Ag (s) + O₂ (g)

We know, molecular mass of Ag₂O= 231.7 g/mol, and atomic mass of Ag= 107.8 g/mol. Given, mass of Ag₂O=5.50 gm. Number moles of Ag₂O= $\frac{5.50}{231.7}$ = 0.0237 moles.

From the balanced chemical reaction we get 2 (two) moles of Ag₂O produces 4 (four) moles of Ag. So, 0.0237 moles of Ag₂O produces $\frac{4X0.0237}{2}$ moles=0.0474 moles of Ag= 0.0474 X 107.8 g of Ag=5.11g Ag.

Therefore, 5.50 g Ag₂O produces 5.11 g of Ag as per the given balanced chemical reaction.

4 0

3 years ago