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2 years ago

If an atom has 13 electrons how many electron shells does it have

Chemistry

1 answer:

svlad2 [7]2 years ago

4 0

Answer:

Explanation:

If an atom has 13 electrons then it belongs to p-block of periodic table.

s level can accommodate 2 electrons.

p level can accommodate 6 electrons.

13 means 1s2 2s2 2p6 3s2 3p1.

As you can see there totally 5 sub-shells.

Total number of shells are 3(1,2,3).

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Which of the following is a true for subtropical jet streams

vodka [1.7K]

Answer: how do we answer when there are no options??

Explanation:

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3 years ago

1. If I have 45 L of He in a balloon at 25 degrees celsius and increase the temperature of the

Greeley [361]

Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).

We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.

Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.

(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).

7 0

3 years ago

What volume of an hcl solution with a ph of 1.3 can be neutralized by one dose of milk of magnesia?

kvv77 [185]

274 mL H3 O+ and fully neutralized

It will take one teaspoon of Mg(OH)2 to completely neutralize 2.00×10^2mL of H3O+.

<h3>What is the purpose of milk of magnesia?</h3>

For a brief period of time, this medicine is used to relieve sporadic constipation.
It is an osmotic laxative, which means that it works by drawing water into the intestines, which aids in causing bowel movement.

<h3>What dosage of milk of magnesia is recommended for constipation?</h3>

Take Milk of Magnesia once day, preferably before bed, in divided doses, or as prescribed by a physician.
suggested dosage: 30 mL to 60 mL for adults and kids 12 years of age and older. 15 mL to 30 mL for children aged 6 to 11 years.

learn more about milk of magnesia here

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the question you are looking for is

People often take milk of magnesia to reduce the discomfort associated with acid stomach or heartburn. The recommended dose is 1 teaspoon, which contains 4.00x 10^{2} mg of Mg(OH)_2. What volume of an HCl solution with a pH of 1.3 can be neutralized by one dose of milk of magnesia? If the stomach contains 2.00x10^{2}mL of pH 1.3 solution, is all the acid neutralized? If not, what fraction is neutralized?

7 0

1 year ago

Use of chemical substance in our daily life their advantages and disadvantages

vekshin1

Answer:

Explanation:

By observing chemical reactions, we are able to understand and explain how the natural world works. Chemical reactions turn food into fuel for your body, make fireworks explode, cause food to change when it is cooked, make soap remove grime, and much more.

3 0

3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago