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2 years ago

Which of these is the final result of secondadry succession?

Chemistry

2 answers:

Grace [21]2 years ago

5 0

Answer:

Explanation:

ADAPTATIOnn but if thhere would be an option o all the above it would be that

gtnhenbr [62]2 years ago

4 0

I believe that the answer is A: adaptation

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Estimate the standard internal energy of formation of liquid methyl acetate (methyl ethanoate, ch3cooch3) at 298 k from its stan

LenKa [72]

solution:

$the reaction for formation of methyl acetate is\\CH_{3}OH+CH_{3}COOH----------->CH_{3}COOCH_{3}+H_{2}O\\standard internal energy \\\delta u=\delta H -\delta(pv)=\delta H -\delta n\times RT\\where \delta n=change in number of moles=0\\\delta u=\delta H\\=-442kj/mol$

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2 years ago

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Which of these elements are you least likely to find in nature as a pure element (that is, not combined with any other elements

Delvig [45]

The answer is sodium (Na)

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3 years ago

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The acid dissociation constant for boric acid, h3bo3, is 5.8 x 10-10. calculate the dissociation constant, kb , of the hydrogen

Dmitriy789 [7]

The answer is 5.8x10^-10. I did the math and checked other sources as well.

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3 years ago

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What mass of iron is produced from adding 80.0 g of iron(II) oxide (71.85 g/mol) to 20.0 g of magnesium metal? FeO (l) + Mg (l)

nadya68 [22]

Answer:

46.0g of Iron are produced

Explanation:

Based on the chemical reaction:

FeO(l) + Mg(l) → Fe(l) + MgO(s)

1 mole of Iron (II) oxide reacts per mole of Mg to produce 1 mole of iron

To solve this question we need to convert each mass of reactant to moles using its respectives molar masses in order to find limitng reactant. Moles of limiting reactant = Moles of iron produced:

Moles FeO (Molar mass: 71.85g/mol):

80.0g * (1mol / 71.85g) = 1.11moles FeO

Moles Mg (Molar mass: 24.305g/mol)

20.0g * (1mol / 24.305g) = 0.823 moles Mg

As moles of Mg < Moles FeO, Mg is limiting reactant and the moles of Fe are 0.823 moles.

The mass of Iron produced is:

0.823 moles Fe * (55.845g/mol) =

46.0g of Iron are produced

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2 years ago

If a 32.4 gram sample of sodium sulfate (Na2SO4) reacts with a 65.3 gram sample of barium chloride (BaCl2) according to the reac

STALIN [3.7K]

Answer: The theoretical yield of barium sulfate is 50.9 grams

Explanation:

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

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

For sodium sulfate:

Given mass of sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

$\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol$

For barium chloride:

Given mass of barium chloride = 65.3 g

Molar mass of barium chloride = 208.23 g/mol

Putting values in equation 1, we get:

$\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol$

The chemical equation for the reaction of barium chloride and sodium sulfate follows:

$Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl$

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of barium chloride

So, 0.228 moles of sodium sulfate will react with = $\frac{1}{1}\times 0.228=0.228mol$ of barium chloride

As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate produces 1 mole of barium sulfate.

So, 0.228 moles of sodium sulfate will produce = $\frac{1}{1}\times 0.228=0.228moles$ of barium sulfate

Now, calculating the mass of barium sulfate from equation 1, we get:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.228 moles

Putting values in equation 1, we get:

$0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g$

Hence, the theoretical yield of barium sulfate is 50.9 grams

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2 years ago