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3 years ago

What is the percent composition of beryllium (Be) and oxide (O) in BeO? Round

Chemistry

1 answer:

Rus_ich [418]3 years ago

6 0

Answer:

$\% Be=36\%\\\\\% O=64\%$

Explanation:

Hello!

In this case, since the percent compositions of Be and O in beryllium oxide are computed as shown below:

$\% Be=\frac{m_{Be}}{M_{BeO}} *100\%\\\\\% O=\frac{m_{O}}{M_{BeO}} *100\%$

Since atomic masses are respectively 9.01 and 16.0 g/mol and the molar mass of the compound is 25.01 g/mol, the resulting percent compositions are:

$\% Be=\frac{9.01}{25.01} *100\%=36\%\\\\\% O=\frac{16.0}{25.01} *100\%=64\%$

Best regards!

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What is the oxidation number of cl− in the hypochlorite ion clo−?

mariarad [96]

Answer : The oxidation number of chlorine (Cl) is, (+1)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

The given compound is, $ClO^-$

Let the oxidation state of 'Cl' be, 'x'

$x+(-2)=-1\\\\x-2=-1\\\\x=+1$

Therefore, the oxidation number of chlorine (Cl) is, (+1)

5 0

3 years ago

What molarity of nitric acid (HNO3) was used if 2.00 L must be used to prepare 4.5 L of a 0.25 M HNO3 solution?

Ymorist [56]

The molarity of (HNO₃) that was used if 2.00 L must be used to prepare 4.5 L of a 0.25M HNO₃ solution is 0.563 M

<u><em>calculation</em></u>

This is calculated usind M₁V₁=M₂V₂ formula

where,

M₁( molarity ₁) = ?

V₁( volume ₁) = 2.00 L

M₁ (molarity ₂) = 0.25M

V₂( volume₂) = 4.5 L

make M₁ the subject of the formula by diving both side of the formula by V₁

M₁ is therefore = M₂V₂/V₁

M₁ =[ (0.25 M x 4.5 L) / 2.00 L ] =0.563 M

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3 years ago

What is the 7th element on the periodic table?

aniked [119]

Nitrogen

Hope this helps!

4 0

3 years ago

Read 2 more answers

A solution was found to have a 15.6 % transmittance at 500 nm, its wavelength of maximum absorption, using a cell with a path le

Gnesinka [82]

For the absorbance of the solution in a 1.00 cm cell at 500 nm is mathematically given as

A’ = 0.16138

<h3>What is the absorbance of the solution in a 1.00 cm cell at 500 nm?</h3>

Absorbance (A) 2 – log (%T) = 2 – log (15.6) = 0.8069

Generally, the equation for the Beer’s law is mathematically given as

A = ε*c*l

0.8069 = ε*c*(5.00 )

ε*c = 0.16138 cm-1

then for when ε*c is constant

l’ = 1.00

A’ = (0.16138 cm-1)*(1.00 cm)

A’ = 0.16138

In conclusion, the absorbance of the solution in a 1.00 cm cell at 500 nm is

A’ = 0.16138