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3 years ago

Predict the products that will be formed by the reaction below. Al(s)+Na2SO4(s)—>

Chemistry

1 answer:

Ostrovityanka [42]3 years ago

3 0

Answer:

Al₂(SO₄)₃ and Na

Explanation:

Al has a charge of +3, Na has a charge of +1 and SO₄ has a charge of -2. Since cations and anions will bond we know that Al will bond with SO₄ leaving Na by itself (since this is a single replacement reaction). When Al bonds with SO₄ it makes aluminum sulfate which is Al₂(SO₄)₃ and Na will be left by itself.

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Hello summer roses~ help me with these questions

Olegator [25]

Answer:

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2 years ago

Compare the modern (electron cloud) model of the atom with Bohr’s atomic model. Which of these statements describe the two model

AfilCa [17]

Answer:

B. Bohr’s model electrons cannot exist between orbits, but in the electron cloud model, the location of the electrons cannot be predicted.

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C. The modern model explains all available data about atoms; Bohr’s model does not.

Explanation:

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6 0

2 years ago

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Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

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

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

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

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

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

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3 years ago

What does variable mean

DedPeter [7]

In an experiement things that are changing are called variables.

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2 years ago

2 PUIS

avanturin [10]

Answer: C the study of living things

Explanation:

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2 years ago

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