All categories

2 years ago

Which has a higher frequency electron or a neutron why ?

1 answer:

5 0

The electron has a higher frequency compared to the neutron. It can be explained by the way an electron orbits the nucleus of an atom.

According to Quantum Mechanics, electrons do not really orbit the nucleus of an atom. In fact, the most tightly bound state, the 1s orbital, has no angular momentum at all. This would be the state with the most "kinetic energy" and yet there is no "orbital" motion at all in this state.

However, there are frequencies associated with each orbital.

You might be interested in

What is the concentration of a solution that has 15.0 g NaCl dissolved to a total of 750 ml?

Oksana_A [137]

Answer: The concentration of solution is 0.342 M

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of solute (Sodium chloride) = 15 g

Molar mass of sodium chloride = 58.5 g/mol

Volume of solution = 750 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{15g\times 1000}{58.5g/mol\times 750mL}\\\\\text{Molarity of solution}=0.342M$

Hence, the concentration of solution is 0.342 M

6 0

2 years ago

We make a basic solution by mixing 50. mL of 0.10 M NaOH and 50. mL of 0.10 M Ca(OH)2. It requires 250 mL of an HCl solution to

andreyandreev [35.5K]

Answer: The correct answer is Option 5.

Explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the NaOH.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL$

Putting all the values in above equation, we get:

$M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base.

We are given:

$n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL$

Putting values in above equation, we get:

$1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M$

Hence, the correct answer is Option 5.

7 0

2 years ago

How to name the compounds in chemistry using roman numerals khan academy?

andrezito [222]

What are the answer choices?

7 0

2 years ago

g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol

Ronch [10]

Answer:

$M=0.0637M$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)$

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

$n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}$

Finally, the resulting molarity in 30.8 mL (0.0308 L):

$M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M$

Regards.

3 0

2 years ago

Corrosive substances are rarely harmful to human skin. True or false?

maxonik [38]

False

The adjective corrosive means it will definitely harm the skin

8 0

2 years ago