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Delvig [45]
2 years ago
5

A metal oxide with the formula mo contains 15.44% oxygen. in the box below, type the symbol for the element represented by m.

Chemistry
2 answers:
lana66690 [7]2 years ago
8 0

<u>Answer:</u> The element represented by M is Strontium.

<u>Explanation:</u>

Let us consider the molar mass of metal be 'x'.

The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol

It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

\frac{15.44}{100}\times (x+16)=16g/mol\\\\(x+16)=\frac{16g/mol\times 100}{15.44}\\\\x=(103.626-16)g/mol\\\\x=87.62g/mol

The metal atom having molar mass as 87.62/mol is Strontium.

Hence, the element represented by M is Strontium.

Simora [160]2 years ago
8 0

The metal present in the given metal oxide is strontium and its atomic symbol is  \boxed{{\text{Sr}}}.

Further explanation:

Mole is the measure of the amount of substance. Mole is the relation between the mass of the substance and molar mass of a substance. It is defined as the mass of a substance in grams divided by its molar mass (g/mol).

The expresion to calculate the number of moles is as follows:

{\text{Number of moles}}\;=\;\dfrac{{{\text{Given mass}}\left( {\text{g}}\right)}}{{{\text{molar mass}}\left({{\text{g/mol}}}\right)}}    ...... (1)

The given metal oxide MO contains 15.44 % oxygen, it means that 100 g of metal oxide MO contains 15.44 g of oxygen. The amount of metal present in 100 g of metal oxide is,

 \begin{aligned\text{Mass of metal}}\left( {\text{g}} \right)&=\left( {{\text{mass of metal oxide}}} \right)-\left( {{\text{mass of oxygen}}}\right)\\&=100{\text{ g}}-{\text{15}}{\text{.44 g}}\\&={\text{84}}{\text{.66 g}}\\\end{aligned}

The number of moles of oxygen in 100 g of metal oxide can be calculated by dividing mass of oxygen with molar mass of oxygen.

\begin{aligned}{\text{Number of moles}}\;&=\;\frac{{{\text{mass}}\left( {\text{g}} \right){\text{ of }}{{\text{O}}_2}}}{{{\text{molar mass}}\left( {{\text{g/mol}}} \right){\text{ of }}{{\text{O}}_2}}}\\&=\frac{{{\text{15}}{\text{.44 g}}}}{{16{\text{ g/mol}}}}\\&=0.965{\text{ mol}}\\\end{aligned}

According to the molecular formula of metal oxide MO, 1 mole of metal oxide contains 1 mole of oxygen and 1 mole of metal. Therefore, the number of moles of metal present in given metal oxide must be 0.965 moles.

The molar mass of metal can be calculated as follow:

\begin{aligned}{\text{Molar mass}}\left({{\text{g/mol}}}\right)&= \;\frac{{{\text{mass}}\left( {\text{g}} \right){\text{ of metal}}}}{{{\text{number of moles}}\left( {{\text{mol}}} \right)\;{\text{of metal}}}}\\&=\frac{{84.66{\text{ g}}}}{{0.965{\text{ mol}}}}\\&=87.7\,{\text{g/mol}}\\\end{aligned}

According to the periodic table metal which has a molar mass approximately 87.7 g is strontium (Sr). Hence the metal oxide is strontium oxide.

Learn more:

1. Determine the number of atoms in a molecule of white phosphorous.:brainly.com/question/9202946

2. Rank the gases in decreasing order of effusion:brainly.com/question/1946297

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Mole concept

Keywords: Metal oxide, formula unit, MO, metal, symbol of metal, M, strontium, Sr, number of moles and strontium oxide.

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1) How many aluminum atoms are there in 3.50 grams of Al2O3?
drek231 [11]

Answer:

Aluminium atoms =  4.13 *10^22 aluminium atoms

The correct answer is E

Explanation:

Step 1: Data given

Mass of Al2O3 = 3.50 grams

Molar mass of Al2O3 = 101.96 g/mol

Number of Avogadro = 6.022 * 10^23 /mol

Step 2: Calculate moles Al2O3

Moles Al2O3 = mass Al2O3 / molar mass Al2O3

Moles Al2O3 = 3.50 grams / 101.96 g/mol

Moles Al2O3 = 0.0343 moles

Step 3: Calculate moles Aluminium

In 1 mol Al2O3 we have 2 moles Al

in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al

Step 4: Calculate aluminium atoms

Aluminium atoms =  moles aluminium * Number of Avogadro

Aluminium atoms =  0.0686 * 6.022 * 10^23

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3 0
2 years ago
A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percent composition, the student measures out 5.8
ra1l [238]

Answer:

<u>1. Net ionic equation:</u>

  • Cl⁻(aq) +  Ag⁺(aq) → AgCl(s)

<u />

<u>2. Volume of 1.0M AgNO₃</u>

  • 41ml

Explanation:

1. Net ionic equation for the reaction of NaCl with AgNO₃.

i) Molecular equation:

It is important to show the phases:

  • (aq) for ions in aqueous solution
  • (s) for solid compounds or elements
  • (g) for gaseous compounds or elements

  • NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

ii) Dissociation reactions:

Determine the ions formed:

  • NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
  • AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)
  • NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

iii) Total ionic equation:

Substitute the aqueous compounds with the ions determined above:

  • Na⁺(aq) + Cl⁻(aq) +  Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) +  Na⁺(aq) + NO₃⁻(aq)

iv) Net ionic equation

Remove the spectator ions:

  • Cl⁻(aq) +  Ag⁺(aq) → AgCl(s) ← answer

2.  How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl

i) Determine the number of moles of AgNO₃

The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl

The number of moles of AgCl is determined using the molar mass:

  • number of moles = mass in grams / molar mass
  • molar mass of AgCl = 143.32g/mol
  • number of moles = 5.84g / (143.32g/mol) = 0.040748 mol

ii) Determine the volume of AgNO₃

  • molarity = number of moles of solute / volume of solution in liters

  • 1.0M = 0.040748mol / V

  • V = 0.040748mol / (1.0M) = 0.040748 liter

  • V = 0.040748liter × 1,000ml / liter = 40.748 ml

Round to two significant figures: 41ml ← answer

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Answer:

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Explanation:

HCl(aq) + Na2SO3(aq) —> NaCl(aq) + H2SO3(aq)

Let us balance the equation. This is illustrated below:

There are 2 atoms of Na on the left side of the equation and 1atom on the right side. It can be balance by putting 2 in front of NaCl as shown below:

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Now, we have 2 atoms of Cl on the right side and 1 atom on the left side. Thus, it can be balance by putting 2 in front of HCl as shown below:

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A careful look at the equation proved that the equation is balanced as the numbers of the different atoms of the element on both side of the equation are the same.

4 0
2 years ago
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