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2 years ago

How many grams of Co3+ are present in 2.34 grams of cobalt(III) nitrite?

Chemistry

1 answer:

Assoli18 [71]2 years ago

4 0

Answer:

$m_{Co^{3+}}=0.563gCo^{3+}$

Explanation:

Hello there!

In this case, since these mole-mass relationships are understood in terms of the moles of the atoms forming the considered compound, we first realize that the chemical formula of the cobalt (III) nitrate is Co(NO₃)₃ whereas there is a 1:1 mole ratio of the cobalt (III) ion (molar mass = 58.93 g/mol) to the entire compound. In such a way, we first compute the moles of the salt (molar mass = 58.93 g/mol) and then apply the aforementioned mole ratio to obtain the grams of the required cation:

$m_{Co^{3+}}=2.34gCo(NO_3)_3*\frac{1molCo(NO_3)_3}{244.95 gCo(NO_3)_3} *\frac{1molCo^{3+}}{1molCo(NO_3)_3} *\frac{58.93gCo^{3+}}{1molCo^{3+}} \\\\m_{Co^{3+}}=0.563gCo^{3+}$

Best regards!

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Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

Setler [38]

The order of frequency is d < a < c < b

$E_{n} = 13.6 * z^{2} / n^{2}$ eV

where z = atomic mass number

n = energy level

For hydrogen z = 1

Therefore, Energy for n = 1

$E_{n} = 13.6 * 1^{2} / 1^{2}$ eV

= -13.6 eV

for n = 2

$E_{n} = 13.6 * 1^{2} / 2^{2}$ eV

= -3.40 eV

for n = 3

$E_{n} = 13.6 * 1^{2} / 3^{2}$ eV

= -1.51 eV

for n = 4

$E_{n} = 13.6 * 1^{2} / 4^{2}$ eV

= -0.85 eV

for n = 5

$E_{n} = 13.6 * 1^{2} / 5^{2}$ eV

= -0.544 eV

n = 2 to n = 4 (absorption)

ΔE = E4 - E2 = -0.85 - (-3.40) = 2.55 eV

n = 2 to n = 1 (emission)

ΔE = E1 - E2 = -13.6 - (-3.40) = -10.2eV

The negative sign indicates that emission will take place.

n = 2 to n = 5 (absorption)

ΔE = E5 - E2 = -0.544 - (-3.40) = 2.856 eV

n = 4 to n = 3 (emission)

ΔE = E3 - E4 = -1.51 - (-0.85) = -0.66 eV

We know that

E = h * υ

Therefore, Energy is proportional to frequency.

So increasing the order of energy is

E4 < E1 < E3 < E2

order of frequency is

d < a < c < b

For more information click on the link below:

brainly.com/question/17058029

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How many moles of iron can be produced from 4.20 moles of iron (III) oxide?

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According to Reference Table V, which is the strongest Brönsted base?

iogann1982 [59]

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Explanation:

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¿Qué puedes hacer para tratar de recuperar la mayor cantidad de agua de lluvia y qué uso doméstico le puedes dar?

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2 years ago

Given that H2(g) + F2(g) - > 2HF(g) delta H rxn = -546.6 kJ 2H2(g)+ O2(g) - > 2H2O(l) delta H rxn = 571.6 kJ

olasank [31]

Answer:

ΔHrxn = -521.6 kJ

Explanation:

To do this, let's write the equations by separate:

H₂ + F₂ -------> 2HF ΔH = -546.6 kJ

2H₂ + O₂ -------> 2H₂O ΔH = -571.6 kJ

For these reactions, we want to get the following reaction:

2F₂ + 2H₂O -------> 4HF + O₂ ΔHrxn = ?

To do this, all we have to do is take the first two reaction and put them in the way to obtain the third reaction. When we look the final reaction we can see that the water is on the reactants, when originally it was on the product, while Florine is doubled. So all we have to do is rewrite the first two reactions, duplicate the first reaction, and reverse the second reaction, and that way we will get the final reaction:

1) (H₂ + F₂ -------> 2HF ) x2 ΔH = -546.6 kJ x 2

2) (2H₂O --------> 2H₂ + O₂) ΔH = -571.6 kJ x -1

---------------------------------------------------------------------------

1) 2H₂ +2F₂ -------> 4HF ΔH = -1093.2 kJ

2) 2H₂O --------> 2H₂ + O₂ ΔH = +571.6 kJ

Now we sum 1) and 2). In this way, hydrogen cancels out and we do the same with the enthalpy:

1) 2H₂ +2F₂ -------> 4HF ΔH = -1093.2 kJ

2) 2H₂O --------> 2H₂ + O₂ ΔH = +571.6 kJ

---------------------------------------------------------------

2F₂ + 2H₂O -------> 4HF + O₂ ΔHrxn = -1093.2 + 571.6 = -521.6 kJ

So the enthalpy of this reaction is

<em>ΔHrxn = -521.6 kJ</em>

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3 years ago