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4 years ago

8

What is an electron that occupies the outermost energy level or shell of an atom called?

1 answer:

natta225 [31]4 years ago

4 0

It's called a valence electron

And the outermost shell would be called a valence shell.

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Select the pair that has the larger atom or ion listed first.

Sedbober [7]

Answer:

Correc option: $Br^- \, , Kr$

Explanation:

size of atom : it says somthing about how many shell present in a particular atom or ion and it can also be evaluated on the basis of radius of atom.

Br^- and Kr has highest number of shell as compared to other group of species .

Na ,S , Mg ,P all are from 3rd period but Kr and Br^- in the 4th period so size of species of this group will more,

Size increases on increasring the shell number

3 0

3 years ago

NARA [144]

We can skip option B and D because NaCl is salt and H₂SO₄ is a strong acid.
Neutralization reactions are those reactions in which acid and base react to form salt and water.
As water being amphoteric in nature can react with HCl as follow,

HCl + H₂O ⇆ H₃O⁺ + OH⁻

In this case no salt is formed, so we can skip this option.

Ammonia being a weak base can abstract proton from HCl as follow,

HCl + NH₃ → NH₄Cl

Ammonium Chloride is a salt. So, among all four options, Option-C is the correct answer.

3 0

3 years ago

Read 2 more answers

Which of the following has helped preserve resources while cutting the demand for energy?

sweet [91]

A is the correct answer

4 0

3 years ago

Read 2 more answers

Where are electrons found in the atom?

RoseWind [281]

In the electron shell

5 0

3 years ago

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

3 0

3 years ago