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PilotLPTM [1.2K]
3 years ago
13

What kind of bond would a metal and a nonmetal typically make?

Chemistry
2 answers:
Delicious77 [7]3 years ago
7 0

The answer is A. An ionic bond.

lana66690 [7]3 years ago
4 0
A) ionic bond.......
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The radioactivity due to carbon-14 measured in a piece of a wooden casket from an ancient burial site was found to produce 20 co
bekas [8.4K]

Answer:

17202.6 years

Explanation:

Activity of the living sample (Ao) = 160 counts per minute

Activity of the wood sample (A) = 20 counts per minute

Half life of carbon-14 = 5730 years

t= age of the artifact

From;

0.693/t1/2= 2.303/t log Ao/A

Then;

0.693/ 5730= 2.303/t log Ao/A

Substituting values;

0.693/5730= 2.303/t log (160/20)

Then we obtain;

1.209×10^-4 = 2.0798/t

t= 2.0798/1.209×10^-4

Thus;

t= 17202.6 years

Therefore the artifact is 17202.6 years old.

3 0
3 years ago
Odor extensive or intensive
slava [35]

Answer:

Odour is intensive

Explanation:

intensive properties are Independent of the amount of matter

6 0
3 years ago
Which of the following reactions have a positive ΔSrxn? Check all that apply.
PolarNik [594]

Answer:

The reactions that have a <em>positive ΔS rxn </em>are the first and the fourth choices:

  • <em>2A(g) + B(s) → 3C(g)</em>

  • <em>2A(g) + 2B(g) → 5C(g)</em>

Explanation:

<em>ΔS rxn </em>is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn  means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

  • <em><u>2A(g) + B(s) → 3C(g)</u></em>

        1)  The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

        2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: <em>ΔS rxn is positive.</em>

  • <em><u>2A(g) + B(g) → C(g)</u></em>

Using the same reasoning, 3 gas molecules in the  reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

  • <u><em>A(g) + B(g) → C(g)</em></u>

Again, 2 gas molecules in the  reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

  • <u><em>2A(g) + 2B(g) → 5C(g)</em></u>

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: <em>ΔS rxn is positive.</em>

6 0
4 years ago
What volume of chlorine gas at 45.3 c is needed to react with 14.2g of sodium to form nacl at 1.72 atm?
kvasek [131]
To determine the volume of chlorine gas that is needed, we need to first know the reaction. It is expressed as follows:

2Na + Cl2 = 2NaCl

Then, from the amount of NaCl to be produced, we calculate the moles of Cl2 needed.

14.2 g NaCl ( 1 mol / 58.44 g ) ( 1 mol Cl2 / 2 NaCl ) = 0.1215 mol Cl2

To determine the volume of the gas, we need an equation that would relate the number of moles to volume. There are a number of equation available but some are complex equations so we assume that it is an ideal gas. We use the equation:

PV = nRT
V = nRT / P
V = 0.1215 mol ( 0.08205 L-atm / mol-K ) (45.3 +273.15 K) / 1.72 atm
V = 1.85 L Cl2 gas
7 0
3 years ago
Read 2 more answers
Question of organic chemustry .The cyclic compound M is heated with dilute hydrochloric acid what is the product?
kotykmax [81]

HYDROLYSING AMIDES

This page describes the hydrolysis of amides under both acidic and alkaline conditions. It also describes the use of alkaline hydrolysis in testing for amides.

The hydrolysis of amides

What is hydrolysis?

Technically, hydrolysis is a reaction with water. That is exactly what happens when amides are hydrolysed in the presence of dilute acids such as dilute hydrochloric acid. The acid acts as a catalyst for the reaction between the amide and water.

The alkaline hydrolysis of amides actually involves reaction with hydroxide ions, but the result is similar enough that it is still classed as hydrolysis.

Hydrolysis under acidic conditions

Taking ethanamide as a typical amide:

If ethanamide is heated with a dilute acid (such as dilute hydrochloric acid), ethanoic acid is formed together with ammonium ions. So, if you were using hydrochloric acid, the final solution would contain ammonium chloride and ethanoic acid.

Hydrolysis under alkaline conditions

Again, taking ethanamide as a typical amide:

If ethanamide is heated with sodium hydroxide solution, ammonia gas is given off and you are left with a solution containing sodium ethanoate.

Using alkaline hydrolysis to test for an amide

If you add sodium hydroxide solution to an unknown organic compound, and it gives off ammonia on heating (but not immediately in the cold), then it is an amide.

You can recognise the ammonia by smell and because it turns red litmus paper blue.

The possible confusion using this test is with ammonium salts. Ammonium salts also produce ammonia with sodium hydroxide solution, but in this case there is always enough ammonia produced in the cold for the smell to be immediately obvious.

6 0
3 years ago
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