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frutty [35]
2 years ago
10

All redox reactions are combustion reactions. True or False?

Chemistry
1 answer:
VladimirAG [237]2 years ago
8 0

Answer:

the answer is false

Explanation:

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Chemical changes are considered unseen because they occur at the atomic level, changing the actual structure of the thing.
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Physical. chemical only applies when it undergoes a reaction
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3 years ago
Calculate the number of milliliters of 0.710 M Ba(OH)2 required to precipitate all of the Mn2+ ions in 161 mL of 0.796 M KMnO4 s
USPshnik [31]

<u>Answer:</u> The volume of barium hydroxide is 183 mL.

<u>Explanation:</u>

To calculate the moles of cadmium nitrate, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}       .....(1)

Molarity of MnSO_4 = 0.796 M

Volume of MnSO_4 = 161 mL = 0.161 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)

By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with = \frac{1}{1}\times 0.13=0.13mol of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So, 0.183L=0.183\times 1000=183mL

Hence, the volume of barium hydroxide is 183 mL.

7 0
3 years ago
Calcula la masa atómica del Hierro y las partículas subatómicas de cada uno de sus isótopos. Fe-54 (5.82%), Fe-56 (91.66%), Fe-5
tino4ka555 [31]

Answer:

La masa atómica del hierro es 55.847 gramos por mol.

Explanation:

Las masas molares de Fe_{54}, Fe_{56}, Fe_{57} y Fe_{58} son 53.940 gramos por mol, 55.935 gramos por mol, 56.935 gramos por mol y 57.933 gramos por mol, respectivamente. La masa atómica del hierro se determina mediante el siguiente promedio ponderado:

M_{Fe} = \frac{5.82}{100}\times \left(53.940\,\frac{g}{mol} \right)+\frac{91.66}{100}\times (55.935\,\frac{g}{mol})+\frac{2.19}{100}\times \left(56.935\,\frac{g}{mol} \right)+\frac{0.33}{100}\times \left(57.933\,\frac{g}{mol} \right)

M_{Fe} = 55.847\,\frac{g}{mol}

La masa atómica del hierro es 55.847 gramos por mol.

8 0
3 years ago
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