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Nikitich [7]
2 years ago
13

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.

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

<u>Sulfur </u>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.

<u>Potassium</u> 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.

<u>Neon</u> 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.

<u>Barium</u> 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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Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

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Three 6−l flasks, fixed with pressure gauges and small valves, each contain 6 g of gas at 276 k. flask a contains ch4, flask b c
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First, please check the missing part in your question in the attachment.

a) So first, the Rank of pressure:

according to this formula PV = nRT and when n = m/Mw

PV = m/Mw * R*T

when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater 

when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g

∴ Pressure :

 (A) > (B) > (c)

B) The rank of average molecular kinetic energy:

when K = 3/2 KB T

when K is the average kinetic energy per molecule of gas 

and KB is Boltzmann's constant

and T is the temperature (K)

So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:

∴ A = B = C

C) the rank of diffusion rate after the valve is opened:

according to this formula:

R2/R1 = √M1/M2

from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,

when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g

∴ the rank of diffusion:

A > B > C

D) The rank of the Total kinetic energy of the molecules:

when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules 

∵ Mw A < Mw B < Mw C 

∴no .of molecules of A > B >C

∴ the rank of total kinetic energy is:

A > B > C

e) the rank of density:

when ρ = m/ v 

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so the density also will be the same, ∴ the rank of the density is:

A = B = C

F) the rank of the collision frequency:

as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.

∴ Collision frequency will only depend on the no.of molecules

we have no.of molecules of A > B > C as Mw A < B < C 

∴the rank of the collision frequency is:

A > B > C 

6 0
2 years ago
What is the full number for 5.7x10^4
dsp73

Answer:

5.7*10^4 is equal to 57,000.

Explanation:

First, we must multiply 10 by its power, 4. That would be 10 4 times.

10*10*10*10 = 10,000.

Then, we multiply it by 5.7.

5.7*10,000 = 57,000.

Regards!

3 0
2 years ago
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