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2 years ago

The formula for rust can be represented by fe2o3 how many moles of fe are present in 25.7 g

1 answer:

7 0

To calculate the number of moles, simply take the ratio of mass and molar mass. The molar mass of Iron (Fe) is equal to 55.85 g/mol.Therefore the total number of moles is:
moles Fe = 25.7 g / (55.85 g / mol)

moles Fe = 0.46 mole

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Which is the electron configuration for bromine?

r-ruslan [8.4K]

Answer:

bromine atomic number =35

1s²2s²2p^6 3s² 3p^6 4s² 3d^10 4p^5.

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3 years ago

A sealed container with liquid water in the bottom was left standing at a constant temperature until the air above the water sur

Julli [10]

Answer:

Saturation.

Explanation:

Hello,

In this case, the statement is accounting for the saturation vapor pressure as it is the pressure of a vapor which is in equilibrium with its liquid, in other words it is the maximum exerted pressure possible by the vapor at a given temperature or just the maximum amount of the vapor, so there is neither no more vapor that could condense nor more liquid that could boil.

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2 years ago

a closed flask of air (0.250 L) contains 5.00 "puffs" of particles. The pressure probe on the flask reads 93 kPa. A student uses

Sergio039 [100]

Answer: New pressure inside the flask would be 148.8 kPa.

Explanation: The combined gas law equation is given by:

$PV=nRT$

As the flask is a closed flask, so the volume remains constant. Temperature is constant also.

So, the relation between pressure and number of moles becomes

$P=n\\or\\\frac{P}{n}=constant$

$\frac{P_1}{n_1}=\frac{P_2}{n_2}$

Initial conditions:

$P_1=93kPa\\n_1=5\text{ puffs}$

Final conditions: When additional 3 puffs of air is added

$P_2=?kPa\\n_2=8\text{ puffs}$

Putting the values, in above equation, we get

$\frac{93}{5}=\frac{P_2}{8}\\P_2=148.8kPa$

3 0

2 years ago

Which letter pattern best represents seafloor spreading? (Hint: Letter "f" represents the mid-

labwork [276]

Answer:

anna ou

Explanation:

6 0

2 years ago

Read 2 more answers

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t

andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

$OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)$

So, moles of product formed are calculated as follows.

$\frac{3}{2} \times 0.17 mol \\= 0.255 mol$

Hence, the given data is as follows.

$n_{1}$ = 0.17 mol, $n_{2}$ = 0.255 mol

$V_{1}$ = 19.5 L, $V_{2} = ?$

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L$

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

8 0

2 years ago