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evablogger [386]
3 years ago
6

How many valence electrons does aluminum (Al) have available for bonding? 1 2 3 4

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
7 0

Answer: three

Explanation: Al has only three elections available for bonding

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Calculate the number of moles<br> 41.0 g of Ca(NO3)2 <br> 1.48 g of Ca(OH)2 <br> 3.40 g of CaSO4
Lemur [1.5K]
First. moles is just a label for a number of things. just like a dozen = 12, a gross = 144, a mole = 6022 with another 20 zeros after the 2

next

moles = mass / molecular weight.

molecular weight = sum of atomic mass from the periodic table

atomic mass MnO2 = atomic mass Mn + 2 x atomic mass O
= 54.94 + 2 x 16 = 86.94 g/mole

so moles MnO2 = 98.0 grams / (86.94 g/mole) = 1.13 moles

notice that I only gave 3 digits? that because of sig figs read the link below if you don't understand....

mw C5H12 = 5 x 12 + 12 x 1 = 72 g/mole

so moles C5H12 = 12.0 g / 72.0 g/mole = 0.167 moles

mw XeF6 = 131.3+ 6 x 19.00 = 245.3

so moles XeF6 = 100 g / 245.3 g/mole = 0.4077 moles

I've also provided a link to a periodic table. if you need atomic weights click on any element and it will give you the
details.
8 0
4 years ago
What volume of 2.0 M HCl in mL will neutralize 25.0 mL of 1.00 M KOH?
MakcuM [25]
CaVa=CbVb
2xV=1X25
V=25/2
V=12.5ML
3 0
3 years ago
Someone please help me figure this out!
Aliun [14]
Hey!!

here is your answer >>>

let us multiply the two numbers above!. we get,

(7.51)(6.50) ( 10) / 1.62 (10 )^-6

(7.51)(6.5)(10)^7/1.62

(751)(650)(10^3) / 1.62

301327.160 (10^3)

301327160 is the number.

Hope my answer helps!
4 0
3 years ago
N2(g) +O2(g) —&gt; N2O5(g)
Alex Ar [27]

Answer:

2N₂(g) +   5O₂(g) ----------> 2N₂O₅(g)

Nitrogen N2(g) get an coefficient of  2.

Explanation:

Reaction Given

                 N₂(g) +O₂(g) —> N₂O₅(g)

Balanced Equation =?

Balance Chemical Equation:

A balanced chemical equation is that in which the number of the reactant atoms equal to the number of the product atom.

For example if the Nitrogen at the reactant side is 3, the number of Nitrogen must be 3 on the product side.

Check the chemical equation and count the number of atoms at the reactant side and product side.

if the number of atoms of the reactants equal to the number of atoms of  products then the reaction is balanced but if not equal then reaction is not balanced

Balancing is an trial and error process to get a balance reaction.

First we count the number of atoms of reactant and product

Number of atoms of Reactant:

N = 2

O = 2

Number of atoms of Product:

N = 2

O = 5

So this is not a balance equation as the atoms of the reactant atom not equal to the product atoms.

Nitrogen atoms are 2 on both reactant and product side

But

Oxygen atoms are 2 on reactant and 5 on product side that is not equal.

Now balance the equation:

                         2N₂(g) +   5O₂(g) ----------> 2N₂O₅(g)

Now again count the number of atoms of reactant and product

Number of atoms of Reactant:

N = 4

O = 10

Number of atoms of Product:

N = 4

O = 10

So now the atoms are equal on both reactant and product side.

By balancing the reaction nitrogen N2(g) get an coefficient of  2.

5 0
4 years ago
[25 pts] When a 14.2g sample of a compound containing only mercury and oxygen is decomposed, 13.2g of Hg is obtained. What is th
Agata [3.3K]

Answer:

% composition of Hg = 93 %

% composition of O = 7 %

Explanation:

Data Given:

mass of total sample = 14.2 g

mass of mercury (Hg) = 13.2 g

percent composition of Hg =

percent composition of O =

Solution:

First find mass of oxygen

As this compound only have two components that is oxygen and mercury

So

upon decomposition it gives 13.2 g mercury

now to find mass of oxygen we have to subtract mass of mercury from total mass of sample

                 Mass of oxygen =  Total mass of sample - mass of Mercury

                 Mass of oxygen =  14.2 g - 13.2 g

                 Mass of oxygen =  1 g

Now to find percent composition of Hg and Oxygen

% composition = component mass / total mass of sample x 100. . . .(1)

For % composition of mercury Hg

Put values in equation 1

             % composition of Hg = 13.2 g / 14.2 g x 100

               % composition of Hg = 93%

For % composition Oxygen (O)

Put values in equation 1

             % composition of O = 1 g / 14.2 g x 100

               % composition of O = 7 %

6 0
3 years ago
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