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4 years ago

6

Consider a transition of the electron in the hydrogen atom from n=5 to n=9. Determine the wavelength of light that is associated

with this transition.

1 answer:

lesantik [10]4 years ago

5 0

Lol i need to answer more and plus n could also = # 4.65

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A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next fe

notsponge [240]

Answer:

the air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.

8 0

3 years ago

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3

AveGali [126]

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$

$\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g$

Theoretical yield of $CO_2$ = 0.572 g

Experimental yield of $CO_2$ = 0.391 g

Now we have to calculate the percent yield of $CO_2$

$\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100$

$\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%$

Therefore, the percent yield of $CO_2$ is, 68.4 %

6 0

3 years ago

N2(g) + 3H2-> 2NH3(g) How many grams of ammonia (NH3) can be produced from the reaction of 28.0 g of N2 and 25.0 g of H2?

Mars2501 [29]

Answer:

34 g of NH₃ were produced in the reaction

Explanation:

N₂ (g) + 3H₂ (g) → 2NH₃ (g)

Moles of N₂ → 28 g / 28 g/m = 1 mol

Moles of H₂ → 25g / 2 g/m = 12.5 moles

Clearly, the limiting is the nitrogen.

1 mol of N₂ produced 2 moles of ammonia

So, If I have 1 mol, I'll produce the same amount

2 moles of NH₃ = Mol . Molar mass

2 m . 17 g/m = 34 g

4 0

3 years ago

How do the 5 branches of chemistry overlap?

mestny [16]

Answer:

Chemistry, Biology, Medicine, Physics, and Geology

7 0

3 years ago

How are covalent molecules named?

vodka [1.7K]

Answer:

B

Explanation:

APEX

5 0

3 years ago