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Fudgin [204]
3 years ago
11

Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 7.80 moles of magnesium perchlorate, Mg(ClO4)2.

Chemistry
1 answer:
Dmitry [639]3 years ago
5 0
 The number of moles of moles of  Magnesium,chlorine and oxygen atoms in 7.80 moles of Mg(ClO4)2 is calculated as below

find the total number of each atom in Mg(ClO4)2

that is mg = 1 atom
        Cl =  1x2 = 2 atoms
        O  = 4  x2 = 8 atoms

then multiply 7.80 moles with total number of each atom , to get the number moles of each atom
that is

Mg = 7.80 x1= 7.80  moles
cl =   7.80  x2=15.6  moles
O =  7.80 x8= 62.4 moles

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What is the definition of a covalent bond?
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What is the volume of the liquid in the this diagram?<br>Would it be 37 mL or 36.5 mL?
vivado [14]

The volume of the liquid in this diagram shown above would be equal to 36.5 mL.

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8 0
2 years ago
What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
photoshop1234 [79]

Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

4Al + 3O2 —> 2Al2O3

From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol

Mass of Al2O3 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

8 0
3 years ago
what volume of hydrogen gas is evolved from a reaction between 0.52 g of Na and water? This gas is collected at 20 C and 745mmHg
guapka [62]
The  volume of  hydrogen  gas  that evolved  is   calculated  as  follows
by  use  of   ideal  gas  equation

that  is  PV = nRT
P=745  mm hg
V= ?
R(gas  constant)= 62.36 L.mm hg/mol.k
T= 20 + 273 = 293 k
n=number  of  moles which is calculated as  follows
find the  moles  of Na  used
= 0.52/23=0.023  moles

write the reacting equation
2Na +2H2O =2NaOH +H2
by  use  of reacting  ratio  between  Na : H2  which  is  2:1  therefore  the mole  of H2 = 0.023/2 =0.0115  moles

by  making  the  volume  the   subject  of  the formula
v=nRT/P
V= (0.0115 x 62.36  x 293) / 745  = 0.283 L


6 0
3 years ago
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