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3 years ago

A piece of magnesium ribbion reacts with oxygen to form magnesium oxide , MgO . What is the percent composition of the compound

Chemistry

1 answer:

boyakko [2]3 years ago

5 0

Answer:

see explanation below

Explanation:

In this case, we have the following reaction:

Mg + 1/2O₂ -------> MgO

Now, according to this reaction we want to know the percent composition of MgO. The problem is not providing the mass of the initial reactants and the product, so we can use the atomic weights of the components, to do this

The molecular weight of Mg is 24.305 g/mol, and O is 15.999 g/mol, so, let's calculate the molar mass of MgO:

MM MgO = 24.305 + 15.999 = 40.304 g/mol

Now with this weight, let's see the percent composition of this compound:

%Mg = 24.305 / 40.304 * 100 = 60.304 %

%O = 15.999 / 40.304 * 100 = 39.696 %

And this would be the percent composition of MgO

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A cylinder is filled with 10.0L of gas and a piston is put into it. The initial pressure of the gas is measured to be 96.0kPa. T

Norma-Jean [14]

Answer:

The final pressure of the gas is:- 21.3 kPa

Explanation:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 10.0 L

V₂ = 45.0 L

P₁ = 96.0 kPa

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${96.0\ kPa}\times {10.0\ L}={P_2}\times {45.0\ L}$

${P_2}=\frac {{96.0}\times {10.0}}{45.0}\ kPa$

${P_2}=21.3\ kPa$

The final pressure of the gas is:- 21.3 kPa

3 0

3 years ago

Using the same sample of gas (P1 = 565 torr , T1 = 27 ∘C ), we wish to change the pressure to 5650 torr with no accompanying cha

Stels [109]

Answer:

2726.85 °C

Explanation:

Given data:

Initial pressure = 565 torr

Initial temperature = 27°C

Final temperature = ?

Final pressure = 5650 torr

Solution:

Initial temperature = 27°C (27+273 = 300 K)

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

T₂ = P₂T₁ /P₁

T₂ = 5650 torr × 300 K / 565 torr

T₂ = 1695000 torr. K /565 torr

T₂ = 3000 K

Kelvin to degree Celsius:

3000 K - 273.15 = 2726.85 °C

4 0

2 years ago

It increases with. And

sp2606 [1]

Yessss sirrrrrrrrrrr

4 0

3 years ago

The pressure in car tires is often measured in pounds per square inch ( lb/in.2 ), with the recommended pressure being in the ra

Goshia [24]

when we convert 32.5 lb/in² to atmosphere, the result obtained is 2.21 atm

<h3>Conversion scale</h3>

14.6959 lb/in² = 1 atm

<h3>Data obtained from the question</h3>

Pressure (in lb/in²) = 32.5 lb/in²
Pressure (in ATM) =?

<h3>How to convert 32.5 lb/in² to atm</h3>

14.6959 lb/in² = 1 atm

Therefore

32.5 lb/in² = 32.5 / 14.6959

32.5 lb/in² = 2.21 atm

Thus, 32.5 lb/in² is equivalent to 2.21 atm

Learn more about conversion:

brainly.com/question/2139943

#SPJ1

4 0

1 year ago

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given

tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

$K_a$ = $1.34*10^{-5$

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝) $\alpha = \sqrt{\frac{K_a}{C} }$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }$

$\alpha = \sqrt{4.9629*10^{-5}}$

$\alpha = 7.044*10^{-3$

percentage% (∝) = $7.044*10^{-3}*100$

= 0.704%

For Part B; where Concentration of B = $7.84*10^{-2$ M

$\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }$

$\alpha = \sqrt{1.709*10^{-4} }$

$\alpha = 0.0130\\$

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= $1.92*10^{-2} M$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }$

$\alpha = \sqrt{6.802*10^{-4}}$

$\alpha =0.02608$

percentage% (∝) = 0.02608 × 100%

= 2.60%

7 0

3 years ago