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2 years ago

Which element has the same number of valence electrons as calcium?

Chemistry

1 answer:

kipiarov [429]2 years ago

8 0

Answer:

Barium has the same number of valence electrons as calcium

Explanation:

Valence electrons is the number of electrons of an atom on the outer shell.

Those valence electrons can participate in the formation of a chemical bond (if the outer shell is not closed); in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair.

Calcium is an atom, part of group 2, called the alkaline earth metals. The alkaline earth metals have 2 valence electrons.

Sulfur is part of a group 16, called the chalcogens or oxygen family. Those atoms have 6 valence electrons. They can form a bound with atoms of group 2 such as calcium, but do not have the same number of valence electrons.

Potassium is part of group 1, called the alkali metals or lithium family. Those atoms have 1 valence electrons. That means Potassium do not have the same number of valence electrons like calcium.

Neon is part of group 18, the noble gasses. Those are stable atoms, which means they have 8 valence electrons. They do not have the same number of valence electrons like Calcium.

Barium an atom, part of group 2, called the alkaline earth metals. The alkaline earth metals have 2 valence electrons. Calcium is also part of this group.

This means barium has the same number of valence electrons as Calcium.

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Draw Au3N as a picture of atoms

Shkiper50 [21]

The picture of Au₃N is attached below.

The first part of the picture shows the formation of Au and N ions.

Formation of Au⁺¹ :

As Gold has one valence electron in 6s¹ therefore, it will loose it to form Au⁺¹. In case of Au₃N three atoms of Au looses three electrons to form three Au⁺¹ ions.

Formation of N⁻³ :

As Nitrogen has 5 valence elctrions therefore, it will gain three electrons that lost by Au to form Nitrite (i.e. N⁻³)

Formation of Au₃N:

Three cations of Au⁺ combines with one anion of N⁻³ to form a neutral ionic compound i.e. Au₃N as shown in second part of the picture.

7 0

3 years ago

When 2.3 × 10^3 g of CaCO3 are heated, the actual yield of CaO is 1.09 × 10^3g. What is

iren [92.7K]

The percent yield : 4. 84.58%

<h3>Further explanation</h3>

Reaction

CaCO₃ ⇄ CaO+CO₂

mass CaCO₃ = 2.3 × 10³ g

mol CaCO₃ (MW=100.0869 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{2.3\times 10^3}{100,0869}\\\\mol=22.98$

From the equation, mol CaCO₃ : CaO = 1 : 1, so mol CaO=22.98

mass CaO(MW=56.0774 g/mol)⇒ (theoretical) :

$\tt mass=mol\times MW\\\\mass=22.98\times 56,0774\\\\mass=1288.659~g$

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{1090}{1288.659}\times 100\%\\\\\5yield=84.58\%$

5 0

3 years ago

A gas under a pressure of 74 mmHg and at a temperature of 75°C occupies a 500.0-L container. How many moles of gas are in the co

Leona [35]

Answer : The number of moles of gas present in container are 1.697 mole.

Explanation :

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of gas = 74 mmHg = 0.097 atm

conversion used : (1 atm = 760 mmHg)

V = volume of gas = 500.0 L

T = temperature of gas = $75^oC=75+273=348K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get the number of moles of gas in the container.

$(0.097atm)\times (500.0L)=n\times (0.0821L.atm/mole.K)\times (348K)$

$n=1.697mole$

Therefore, the number of moles of gas present in container are 1.697 mole.

4 0

3 years ago

If 1.80 moles of NaCl was dissolved in enough water to make 3.60 L of solution, what is the Molarity?

sweet-ann [11.9K]

Answer:

0.5M

Explanation:

The equation for molarity is:

M = $\frac{mol}{liters}$ ; where the "M" stands for molarity, the "mol" stands for moles of solute and the "liters" means the volume in liters of solution.

We are given that there are:

1.80 moles of NaCl (the moles of solute)

3.60 Liters of solution (the volume in liters of solution)

Now we just plug those numbers into the formula and get our answer:

M= $\frac{1.80mol}{3.60L}$ = 0.5M

After doing the math and dividing the moles of solute by the liters of solution, we get that the molarity of the solution is 0.5M.

8 0

3 years ago

A 20.0 milliliter sample of 0.200 molar K₂CO₃ solution is added to 30.0 milliliters of 0.400 molar Ba(NO₃)₂ solution. Barium car

sveticcg [70]

Answer:

0.32M

Explanation:

Step 1: Balance the reaction

K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3

We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution

SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3

Step 2: Calculate concentration

To find the concentration of the barium cation we use the following equation:

Concentration = moles of the solute / volumen of the solution

[Ba2+] = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3

[Ba2+] = 0.32 M

The concentration of Barium ion in solution is 0.32 M

6 0

2 years ago