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2 years ago

Use the periodic table to classify each of the elements below.

Chemistry

2 answers:

Alika [10]2 years ago

5 0

Answer:

Explanation:

Cadmium:(Cd)

Cadmium is transition metal present in group twelve. It is soft metal and properties are similar to the other group members like zinc and mercury. Its atomic number is forty eight and have two valance electrons.

Electronic configuration:

Cd₄₈ = [Kr] 4d¹⁰ 5s²

Vanadium: (V)

It is present in group five. It is malleable and ductile transition metal. Its atomic number is twenty three. Vanadium have five valance electrons.

Electronic configuration:

V₂₃ =[Ar] 3d³ 4s²

Xenon: Xe

Xenon is present in group eighteen. It is noble gas. Its outer most valance shell is complete that's why it is inert. its atomic number is fifty four. Xenon have eight valance electrons.

Electronic configuration:

Xe₅₄ = [Kr] 4d¹⁰ 5s² 5p⁶

Iodine: (I)

Iodine is present in group seventeen. Its outer most valance shell have seven electrons. Iodine is the member of halogen family. It gain one electron to complete the octet. its atomic number is fifty three.

Electronic configuration:

I₅₃ = [Kr] 4d¹⁰ 5s² 5p⁵

Potassium: (K)

Potassium is present in group one. it is alkali metal. Its atomic number is nineteen. Its valance shell has one electron. Potassium loses its one valance electron and gets stable electronic configuration.

Electronic configuration:

K₁₉ = [Ar] 4s¹

Strontium: Sr

Strontium is present in group two. it is alkaline earth metal. its atomic number is thirty eight and have two valance electrons.

Electronic configuration:

Sr₃₈ = [Kr] 5s²

dalvyx [7]2 years ago

5 0

Answer:

cadmium (Cd): transition metal

vanadium (V): transition metal

xenon (Xe): noble gas

iodine (I): halogen

potassium (K): alkali metal

strontium (Sr): alkaline earth metal

Explanation:

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When C2H5Cl(g) is burned in oxygen, chlorine gas is produced in addition to carbon dioxide and water vapor. 5145 kJ of heat are

Arisa [49]

<u>Answer:</u> The chemical equation is written below.

<u>Explanation:</u>

Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

$\text{hydrocarbon}+O_2\rightarrow CO_2+H_2O$

The chemical equation for the combustion of ethyl chloride follows:

$4C_2H_5Cl+13O_2\rightarrow 2Cl_2+8CO_2+10H_2O$

We are given:

When 4 moles of ethyl chloride is burnt, 5145 kJ of heat is released.

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

$A+\text{heat}\rightleftharpoons B$

For an exothermic reaction, heat is getting released during a chemical reaction and is written on the product side

$A\rightleftharpoons B+\text{heat}$

So, the chemical equation follows:

$4C_2H_5Cl+13O_2\rightarrow 2Cl_2+8CO_2+10H_2O+5145kJ$

Hence, the chemical equation is written above.

7 0

2 years ago

Lithium has an atomic mass of 6.941 amu. Lithium has two common isotopes. The one isotope has a mass of 6.015 amu and a relative

koban [17]

Answer:

The atomic mass of second isotope is 7.016

Explanation:

Given data:

Average Atomic mass of lithium = 6.941 amu

Atomic mass of first isotope = 6.015 amu

Relative abundance of first isotope = 7.49%

Abundance of second isotope = ?

Atomic mass of other isotope = ?

Solution:

Total abundance = 100%

100 - 7.49 = 92.51%

percentage abundance of second isotope = 92.51%

Now we will calculate the mass if second isotope.

Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

6.941 = (6.015×7.49)+(x×92.51) /100

6.941 = 45.05235 + (x92.51) / 100

6.941×100 = 45.05235 + (x92.51)

694.1 - 45.05235 = (x92.51)

649.04765 = x 92.51

x = 485.583 /92.51

x = 7.016

The atomic mass of second isotope is 7.016

3 0

2 years ago

Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Exp

weeeeeb [17]

Answer : The grams of $N_2$ consumed is, 89.6 grams.

Solution : Given,

Mass of $CaCN_2$ = 265 g

Molar mass of $CaCN_2$ = 80 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate the moles of $CaCN_2$ .

$\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles$

The given balanced reaction is,

$CaC_2+N_2\rightarrow CaCN_2+C$

from the reaction, we conclude that

As, 1 mole of $CaCN_2$ produces from 1 mole of $N_2$

So, 3.2 moles of $CaCN_2$ produces from 3.2 moles of $N_2$

Now we have to calculate the mass of $N_2$

$\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2$

$\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g$

Therefore, the grams of $N_2$ consumed is, 89.6 grams.

5 0

2 years ago

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Sveta_85 [38]

Answer: C
I hope this helped you

6 0

2 years ago

During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.

Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.

it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of CH₄ needs → 2 mol of O₂

??? mol of CH₄ needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over * molar mass

= 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0

2 years ago