All categories

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

You might be interested in

How many moles of ScCl3 can be produced when 10.00 mol Sc react with 9.00 mol Cl2

Nuetrik [128]

Answer:

Moles of $ScCl_3$ = 6 moles

Explanation:

The reaction of $Sc$ and $Cl_2$ to make $ScCl_3$ is:

$2Sc+3Cl_2$ ⇒ $2ScCl_3$

The above reaction shows that 2 moles of Sc can react with 3 moles of $Cl_2$ to form $ScCl_3.$

Mole Ratio= 2:3

For 10 moles of Sc we need:

Moles of $Cl_2$ = $Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ = $10 *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ =15 moles

So 15 moles of $Cl_2$ are required to react with 10 moles of $Sc$ but we have 9 moles of $Cl_2$ , it means $Cl_2$ is limiting reactant.

$Moles of ScCl_3=Given\ Moles\ of\ Cl_2 *\frac{2\ Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}$

$Moles\ of\ ScCl_3=9 *\frac{2\ Moles\ of\ ScCl_3}{3\ Moles\ of\ Cl_2}$

Moles of ScCl_3= 6 moles

4 0

3 years ago

Read 2 more answers

scandium47 has a half-life of 35s. suppose you have a 45g sample of scadium 47 how much of the sample remains unchanged after 14

Mars2501 [29]

The much of the sample that would remain unchanged after 140 seconds is 2.813 g

Explanation

Half life is time taken for the quantity to reduce to half its original value.

if the half life for Scandium is 35 sec, then the number of half life in 140 seconds

=140 sec/ 35 s = 4 half life

Therefore 45 g after first half life = 45 x1/2 =22.5 g

22.5 g after second half life = 22.5 x 1/2 =11.25 g

11.25 g after third half life = 11.25 x 1/2 = 5.625 g

5.625 after fourth half life = 5.625 x 1/2 = 2.813

therefore 2.813 g of Scandium 47 remains unchanged.

4 0

3 years ago

Observe as equações e de acordo com Brönsted-Lowry, os compostos destacados são, respectivamente

Alex777 [14]

answer

j'ai besoin d'aide

7 0

3 years ago

Draw the product of the reaction between CH3CH=CHCH3 and H2 under a platinum catalyst.

Pani-rosa [81]

Answer:

CH₃CH₂-CH₂CH₃

Explanation:

When an alkene, R-CH=CH-R reacts with H₂ in a Pt catalyst, the analogue alkane, R-CH₂-CH₂-R, is produced (Hydrogenation of alkenes via Pt/Pd catalyst)

Thus, the reaction of CH₃CH=CHCH₃ with H₂ under a platinum catalyst produce:

The analogue alkane

4 0

3 years ago

A researcher studying the nutritional value of a new candy places a 4.90 g sample of the candy inside a bomb calorimeter and com

snow_lady [41]

Answer:

449730.879 cal/g

Explanation:

Given data:

Mass of sample = 4.9 g

Change in temperature = 2.08 °C (275.23 k)

Heat capacity of calorimeter = 33.50 KJ . K⁻¹

Solution:

C(candy) = Q/m

Q = C (calorimeter) × ΔT

C(candy) = C (calorimeter) × ΔT / m

C(candy) = 33.50 KJ . K⁻¹ × 275.23 K / 4.90 g

C(candy) = 9220.205 KJ / 4.90 g

C(candy) = 1881.674 KJ / g

It is known that,

1 KJ /g = 239.006 cal/g

1881.674 × 239.006 = 449730.879 cal/g

8 0

3 years ago