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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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The entropy change for a real, irreversible process is equal to:______.

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Answer:

The entropy change for a real, irreversible process is equal to zero.

The correct option is 'c'.

Explanation:

Lets look around all the given options -:

(a) the entropy change for a theoretical reversible process with the same initial and final states , since the entropy change is equal and opposite in reversible process , thus this option in not correct.

(b) equal to the entropy change for the same process performed reversibly ONLY if the process can be reversed at all. Since , the change is same as well as opposite too . Therefore , this statement is also not true .

(c) zero. This option is true because We generate more entropy in an irreversible process. Because no heat moves into or out of the surroundings during the procedure, the entropy change of the surroundings is zero.

(d) impossible to tell. This option is invalid , thus incorrect .

Hence , the correct option is 'c' that is zero.

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There are 3 different possible structures (known as isomers) for a dibromoethene molecule, C2H2Br2. One of them has no net dipol

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Explanation:

Compounds having same molecular formula but different structural and spatial arrangement are isomers.

Three isomers are possible for dibromomethene.

In one structure (IUPAC name: 1,1-dibromomethene), both the bromine atoms are attached to one carbon atom.

In another two structures (Cis and trans), two bromine atoms are attached to two different carbon atoms.

In Cis 1,2-dibromomethene, two bromine atoms are present on the same side.

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3 years ago

A. Calculations for the Determination of Ammonium Chloride The data from the data entry portion of the report has been copied in

Vladimir79 [104]

Answer:

Explanation:

Considering question A

Mass of original sample is $m_o = 0.945 \ g$

Mass of NH4Cl is $m_n = 0.116 \ g$

Percent of NH4Cl is $k = 12.275 \%$

Mass of NaCl $m_k = 0.359 \ g$

Mass of SiO2 $m_e = 0.46$

Mass of original sample $m_o = 0.945 \ g$

Differences in these weights (g) (use the absolute value of the difference)

recovery of matter $G = 0.01 \ g$

The correct option is C

From the question we are told that

The mass of evaporating dish on #1 is $m_1 = 38.646 \ g$

The mass of evaporating dish and original sample $m_2 = 39 591 \ g$

The mass of evaporating dish after subliming $NH_4Cl$ is $m_3 = 39.4750 \ g$

Generally the mass of the original sample is mathematically represented as

$m_o = m_2 - m_1$

=> $m_o = 39 591 - 38.646$

=> $m_o = 0.945 \ g$

Generally the mass of $NH_4Cl$ is mathematically represented as

$m_n = m_2 - m_3$

=> $m_n = 39 591 - 39.4750$

=> $m_n = 0.116 \ g$

The Percent $NaH_4 Cl (g)$

$k = \frac{ m_n}{m_o} *100$

=> $k = \frac{0.116 }{0.945} *100$

=> $k = 12.275 \%$

Considering question B

The mass of evaporating dish #2 is $m_g = 38700\ g$

The mass of watch glass is $m_a = 28 299 \ g$

The mass of evaporating dish #2, watch glass and NaCl $m_b = 67,355 \ g$

Generally the mass of NaCl is

$m_k = m_b -[m_g + m_a]$

=> $m_k = 67,355 -[38700 + 28 299]$

=> $m_k = 0.359 \ g$

Considering question C

The mass of evaporating dish is $m_p= 38.645$

The mass of evaporating dish and SiO2 $m_s = 39.105 \ g$

Generally the mass of SiO2 is mathematically represented as

$m_e = 39.105 - 38.645$

=> $m_e = 0.46$

Considering D

The mass of the original sample is $m_o = 0.945 \ g$

Generally the experimental mass recovered (NH_4Cl,NaCl, SiO2 ) is mathematically evaluated as

$M =0.116 + 0.46 + 0.359$

$M = 0.935 \ g$

Generally the differences in these weights (g) of recovery of matter is mathematically represented as.

$G =0.945- 0.935$

=> $G = 0.01 \ g$

While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the recovered mass to be lower.

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3 years ago