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3 years ago

How many electrons are contained in a co3+ ion

1 answer:

7 0

The atomic number of cobalt, CO is 27. Since, the cobalt that you asked is in ion. The number of electrons should be 27-3 --> *24* Reasons being that cobalt has donated/given away 3 electrons.

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It might be to long but please help. I did search but couldn't find answer

Inessa05 [86]

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3 years ago

When warm air rises and cool air sinks towards earth surface. what is this known as?

iVinArrow [24]

When warm air rises and cool air sinks towards earth surface. what is this known as convection currents.

Explanation:

Convection currents not only occurs in the air but also in other fluids like water. In the air, they are responsible for the blowing of winds of the atmosphere . In the oceans, they are responsible for ocean currents. Convection currents also occur in the mantle and are responsible for tectonic plate movements.

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4 0

4 years ago

Consider the balanced equation for the following reaction:

Zanzabum

Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol$

For the given chemical equation:

$7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)$

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = $\frac{4}{7}\times 0.25=0.143mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g$

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

7 0

3 years ago

Will give brainliest

zhenek [66]

Answer:

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8 0

3 years ago

If a temperature increase from 12.0 ∘c to 22.0 ∘c doubles the rate constant for a reaction, what is the value of the activation

Alex73 [517]

Answer: 48,501 J/mol

Explanation:

1) Action barrier = activation energy = Ea

2) Data:

i) T₁ = 12°C = 12 + 273.15 K = 285.15K

ii) T₂ = 22°C = 22 + 273.15 K = 295.15 K

iii) rate constant = k: k₂ / k₁ = 2

iv) Ea = ?

3) Formula:

Arrhenius' law gives the relationship between the constant of reaction and the temperature:

$k=Ae^{-\frac{Ea}{RT}}$

4) Solution

By arranging the formula, you get:

㏑[k₂/k₁] =Ea/R [1/T₁ - 1/T₂]

Replace k₂ = 2k₁; T₁ = 285.15; and T₂ = 295.15

ln[2] = Ea/8.314 J/K mol × [1/285.15 - 1/295.15]K

Ea = ln [2] × 8.314 J/K mol / [1.18818×10⁻⁴K] = 48,501 J/mol

6 0

4 years ago