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Brrunno [24]
3 years ago
14

Find the % composition for each element in Zn(CIO3)2

Chemistry
1 answer:
bezimeni [28]3 years ago
7 0

Zn = 28.15%

Cl = 30.53%

O = 41.32%

<h3>Further explanation</h3>

Given

Zn(CIO3)2 compound

Required

The % composition

Solution

Ar Zn = 65.38

Ar Cl = 35,453

Ar O = 15,999

MW Zn(CIO3)2 = 232.3

Zn = 65,38/232.3 x 100% = 28.15%

Cl = (2 x 35.453) / 232.3 x 100% = 30.53%

O = (6 x 15.999) / 232.3 x 100% = 41.32%

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Write the balanced COMPLETE ionic equation for the reaction when
Vera_Pavlovna [14]

Answer:

Ba²⁺(aq) + 2 NO₃⁻(aq) + 2 Rb⁺(aq) + 2 OH⁻(aq) = Ba(OH)₂(s) + 2 Rb⁺(aq) + 2NO₃⁻(aq)

Explanation:

Let's consider the molecular equation between barium nitrate and rubidium hydroxide to produce barium hydroxide and rubidium nitrate.

Ba(NO₃)₂(aq) + 2 RbOH(aq) = Ba(OH)₂(s) + 2 RbNO₃(aq)

The complete ionic equation includes all the ions and the molecular species.

Ba²⁺(aq) + 2 NO₃⁻(aq) + 2 Rb⁺(aq) + 2 OH⁻(aq) = Ba(OH)₂(s) + 2 Rb⁺(aq) + 2NO₃⁻(aq)

3 0
3 years ago
How can you determine the specific heat of a metal using a calorimeter
zhannawk [14.2K]

Answer:

One can determine the specific heat of the metal through using the clarimeter, water, thermometer and using heat equations.

Explanation:

You can learn about heat effects and calorimetery through a simple experiment by boiling water and heating up the metal in it. Then, pour it into your calorimeter and the heat will flow from the metal to the water. The two equlibria will meet: the metal will loose heat into its surroundings (the water) and teh water will absorb the heat. The heat flow for the water is the same as it is for the metal, the only difference being is the negative sign indicating the loss of the heat of the metal.

In terms of theromdynamics, we can deteremine the heat flow for the metal becasue it would be equal to the mangnitued but opposite in direction. Thus, we can say that the specific heat of water qH2O = -qmetal.

4 0
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What mass of octane must be burned in order to liberate 5270 kj of heat? δhcomb = -5471 kj/mol?
katrin2010 [14]
As,
                           5471 kJ heat is given by  =  1 mole of Octane
Then,
                    5310 kJ heat will be given by  = X moles of Octane

Solving for X,
                                  X  =  (5310 kJ × 1 mol) ÷ 5471 kJ

                                  X  =  0.970 moles of Ocatne

So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
                                  Moles  =  mass / M.mass
Or,
                                  Mass  =  Moles × M.mass
Putting values,
                                  Mass  =  0.970 mol × 114.23 g/mol

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6 0
3 years ago
Given 4.80g of ammonium carbonate, find:
V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

<em>1) Number of moles of the compound:</em>

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

<em>2) Number of moles of ammonium ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) =  <em>0.1 mol.</em>

<em>3) Number of moles of carbonate ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

<em>4) Number of moles of hydrogen atoms:</em>

  • Every 1.0 mol of (NH₄)₂CO₃  contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>

<em>5) Number of hydrogen atoms:</em>

  • It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

<u><em>Using cross multiplication:</em></u>

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

<em>∴ The no. of atoms in  0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>

8 0
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