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3 years ago

12

A lamp releases only light and thermal energy. The energy input of the lamp is 120 watts. Of that input, 80 watts is converted t

o light. How much energy is converted to thermal energy?

1 answer:

fomenos3 years ago

6 0

40 watts because 120-80 is forty or 80 + 40 - 120

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2. Fe(s) + O2(g) → Fe3O4(s) a. When 13.54 g of O2 is mixed with 12.21 g of Fe, which is the limiting reactant? b. What mass in g

julsineya [31]

The percent yield shows the extent to which the reactants are converetd into products . The limiting reactant is used up in the reaction.

<h3>What is a limiting reactant?</h3>

A limiting reactant is the reactant that is in the least amount in the system. Now;

Number of moles of Fe = 12.21 g/56 g/mol = 0.22 moles

Number of moles of O2 = 13.54 g/32 g/mol = 0.42 moles

Balanced reaction equation;

3Fe(s) + 2 O2(g) = Fe3O4(s)

If 3 moles of Fe reacts with 2 mole of O2

0.22 moles of Fe reacts with 0.22 moles * 2 mole/3 moles = 0.15 moles

Hence, Fe is the limiting reactant

If 3 mole of Fe produces 1 mole of Fe3O4(s)

0.22 moles of O2 produces 0.22 moles * 1 mole/3 moles of Fe3O4(s) = 0.1073 moles

Mass of Fe3O4(s) =0.1073 moles * 232 g/mol =16.9 g

Number of moles of excess reactant = 0.42 moles - 0.15 moles = 0.27 moles

Mass of excess reactant = 0.27 moles * 32 g/mol = 8.64 g

percent yield = 15.88 g/16.9 g * 100/1

= 93.4%

Learn more about percent yiled: brainly.com/question/13463225

8 0

2 years ago

Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc

max2010maxim [7]

<u>Answer:</u> The percent yield of the water is 31.98 %

<u>Explanation:</u>

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

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

<u>For methane:</u>

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol$

<u>For oxygen gas:</u>

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol$

The chemical equation for the combustion of methane is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = $\frac{1}{2}\times 0.45=0.225mol$ of methane

As, given amount of methane 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

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = $\frac{2}{2}\times 0.45=0.45$ moles of water

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

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

$0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g$

To calculate the percentage yield of water, we use the equation:

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

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%$

Hence, the percent yield of the water is 31.98 %

4 0

3 years ago

During the experiment a student precipitated and digested the BaSO4. After allowing the precipitate to settle, they added a few

OlgaM077 [116]

Answer:

Incomplete precipitation of barium sulfate

Explanation:

The student has precipitated and digested the barium sulfate on his/her side. But on the addition of $BaCl_2$ in the solution, the solution become cloudy. This happened because incomplete precipitation of barium sulfate by the student. When $BaCl_2$ is added, there are still sulfate ions present in the solution with combines with $BaCl_2$ and forms $BaSO_4$ and the formation of this precipitate makes the solution cloudy.

7 0

3 years ago

How to calculate work done with change in volume and pressure.

natali 33 [55]

$W = \displaystyle \int dW = \displaystyle \int^{V_2}_{V_1} p ~dV = p(V_2 - V_1) = p \Delta V$

3 0

2 years ago

When steam turns a turbine, what kind of energy is being manifest?

mihalych1998 [28]

Answer: mechanical

Explanation:

its a big mechanicale thing

6 0

3 years ago

Read 2 more answers