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3 years ago

PLEASE HELP In the following atomic model, where does the strong nuclear force happen?

Chemistry

2 answers:

Lisa [10]3 years ago

7 0

The answer would be the closest to the inside, so the last option. Hope I helped :)

icang [17]3 years ago

7 0

Answer: Option (D) is the correct answer.

Explanation:

Strong nuclear force is defined as the force that is able to hold both the protons and neutrons together in an atomic nucleus.

Since protons have a positive charge and neutrons have no charge. And like charges repel each other. Therefore, protons repel each other due to electrostatic force of repulsion. This force of repulsion is overcome by nuclear binding energy that is also known as strong nuclear force.

This binding energy helps the protons to be bind with each other.

Thus, we can conclude that strong nuclear force will happen inside C.

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NEED HELP ASAP

hammer [34]

<u>The answer is B. Gas molecules are in Constant Motion.</u>

This is supported by the Kinetic Theory of Gases, which it states that the gas particles are in constant random motion, colliding with the container that it's in.

7 0

3 years ago

Read 2 more answers

If 25.16 g of chlorine react with 12.99 g of manganese metal, what is the empirical formula of the compound?

WARRIOR [948]

We use the given masses of the reactants to calculate the moles of Mn and Cl. Empirical formula represents the simplest mole ratio of atoms present in a compound.

Moles of Mn = $12.99 g Mn * \frac{1 mol Mn}{54.94 g Mn} = 0.236 mol Mn$

Moles of Cl = $25.16 g Cl_{2} *\frac{1 mol Cl_{2}}{70.91 g Cl_{2}} * \frac{2 mol Cl}{1 mol Cl_{2}}$ = 0.710 mol Cl

Simplest mole ratio: $Mn_{\frac{0.236}{0.236}}Cl_{\frac{0.710}{0.236}}$

So the empirical formula is $MnCl_{3}$

6 0

3 years ago

Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H

solmaris [256]

Answer:

36.37% is the percent yield of the reaction.

Explanation:

$4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)$

1)0.650 L nitrogen gas , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T = 295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol$

2) Moles of ammonia gas= $\frac{2.53 g}{17 g/mol}=0.1488 mol$

Moles of oxygen gas = $\frac{3.53 g}{32 g/mol}=0.1101 mol$

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

$\frac{4}{3}\times 0.1101 mol=0.1468 mol$ of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

$\frac{2}{3}\times 0.1101 mol=0.0734 mol$ of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

$\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the reaction:

$\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%$

36.37% is the percent yield of the reaction.

3 0

2 years ago

How were atomic models developed when no one had seen an atom?

AnnyKZ [126]

Atomic models were developed through indirect observation even though no one had seen an atom. There were many experiments of which I know nothing about, but in the end the scientists managed to come up with a formula and various models to describe atoms.

4 0

3 years ago

In calculating the equilibrium constant for a reaction, the coefficients of the chemical equation are used as exponents for the

BartSMP [9]

Answer: The given statement is TRUE.

Explanation:

An equilibrium reaction is one in which rate of forward reaction is equal to the rate of backward reaction.

Equilibrium constant is defined as the ratio of the product of the concentration of products to the product of the concentration of reactants each raised to their stochiometric coefficient.

For example for the given equilibrium reaction;

$2H_2O(g)\leftrightharpoons 2H_2(g)+O_2(g)$

$K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}$

Thus the given statement that in calculating the equilibrium constant for a reaction, the coefficients of the chemical equation are used as exponents for the factors in the equilibrium expression is True.

7 0

3 years ago