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2 years ago

5) When heated in a flame, the element Indium emits electromagnetic radiation with a distinctive indigo blue

Chemistry

1 answer:

grandymaker [24]2 years ago

5 0

Answer:

$\lambda=451nm$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the following equation, defined in terms of energy, Planck's constant, wavelength and speed of light:

$E=\frac{hC}{\lambda }$

Thus, we solve for the wavelength as shown below:

$\lambda=\frac{hC}{ E}$

And finally plug in the energy, Planck's constant and speed of light to obtain:

$\lambda=\frac{6.6261 x 10^{-34} m^2 kg / s*3x10^8m/s}{4.405x10^{-19}m^2kg/s^2}\\\\\lambda=4.513x10^{-7}m*\frac{1nm}{10^{-9}m} \\\\\lambda=451nm$

Regards!

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Determine how many grams of silver would be produced, if 12.83 x 10^23 atoms of copper react with an excess of silver nitrate. G

AnnyKZ [126]

1) Chemical equation

Cu + 2AgNO3 ---> Cu (NO3)2 + 2Ag

2) molar ratios

1 mol Cu: 2 moles AgNO3 : 1 mol Cu (NO3)2 : 2 mol Ag

3) Convert 12. 83 * 10^23 atoms of Cu in moles

12.83 * 10 ^ 23 atoms / (6.02 * 10^23 atoms / mol) = 2.131 mol Cu

4) Use the proportions

2.131 mol Cu * 2 mol Ag / 1 mol Cu = 4.262 mol Ag

5) Use the atomic mass of silver to convert 4.262 mol in grams

mass = number of moles * atomic mass = 4.262 mol * 107.9 g / mol = 459.9 grams

Answer: 459.9 g

5 0

2 years ago

in collecting the precipitate, why would it be inappropriate to heat the reacted mixture and evaporate off the water?

djverab [1.8K]

In collecting the precipitate, it is inappropriate to heat the reacted mixture and evaporate off the water because it is possible that the mixture contains other substances that precipitates as well when the mixture is being heated so you will not be able to collect what you want.

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2 years ago

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

shutvik [7]

Answer:

79%

Explanation:

1) Balanced chemical equation:

2S + 3O₂ → 2SO₃

2) Mole ratio:

2 mol S : 3 mol O₂ : 2 mol SO₃

3) Limiting reactant:

Number of moles of O₂

n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

Number of moles of S:

n = 7.0 g / 32.065 g/mol = 0.2183 mol S

Ratios:

Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

4) Calcuate theoretical yield (using the limiting reactant):

0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃

5) Yield in grams:

mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g

6) Percent yield:

Percent yield, % = (actual yield / theoretical yield) × 100

% = (7.9 g / 10.0 g) × 100 = 79%

6 0

2 years ago

What volume (ml) of a 0.2450 m koh(aq) solution is required to completely neutralize 55.25 ml of a 0.5440 m h3po4(aq) solution?

Nonamiya [84]

Answer: It depends on what came after "0.5440 M H...". If it was a monoprotic acid, like HCl, the calculation would go like this: (55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) = 122.7 mL KOH If it was a diprotic acid, like H2SO4, like this: (55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) = 245.4 mL KOH If it was a triprotic acid, like H3PO4, like this: (55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) = 368.0 mL KOH

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2 years ago

Rank the formation of the solutions A, B, and C from the most exothermic to the most endothermic. Rank the enthalpy of solution

Elodia [21]

This is an incomplete question, the table is attached below.

Answer : The correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.

Explanation :

As we know that the intermolecular force of attraction play an important role in the interaction of solute-solute, solute-solvent and solvent solvent solution.

In the solution A, the solute-solute and solvent-solvent interactions are weak. So, their solute-solvent interaction will be strong. That means, the solution will be more exothermic.

In the solution C, the solute-solute and solvent-solvent interactions are strong. So, their solute-solvent interaction will be weak. That means, the solution will be more endothermic.

Thus, the correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.

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