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3 years ago

What is the mass of 0.7891 mol of ferric oxide (Fe,O3)?

Chemistry

2 answers:

Natasha2012 [34]3 years ago

4 0

The answer i believe is answer A

Aleksandr [31]3 years ago

3 0

Answer:

The answer to your question is letter B.

Explanation:

Process

1.- Calculate the molecular mass of Ferric oxide

Fe = 56g x 2 = 112 g

O = 16 x 3 = 48 g

Mass = 112 + 48 = 160 g

2.- Calculate the mass of 0.7891 moles of Ferric oxide

160 g of Fe₂O₃ ------------- 1 mol

x ------------- 0.7891 mol

x = (160 x 0.7891) / 1

x = 126.2 g

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Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

For process 2 :

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution

vfiekz [6]

Answer:

450. g of 0.173 % KCN solution contains 779 mg of KCN.

Explanation:

Mass of the solution = m

Mass of the KCN in solution = 779 mg

Mass by mass percentage of KCN solution = 0.173%

$(m/m)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$0.173\%=\frac{779 mg}{m}\times 100$

$m=\frac{779 mg}{0.173}\times 100= 450,289 mg$

1 mg = 0.001 g

m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g

450. g of 0.173 % KCN solution contains 779 mg of KCN.

6 0

3 years ago

Butane, C4H10, reacts with oxygen, O2, to form water, H2O, and carbon dioxide, CO2, as shown in the following chemical equation:

Wittaler [7]

The balanced chemical reaction is:

2C4H10(g)+13O2(g)->10H2O(g)+8CO2(g)

Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
1.77 g C4H10 (1 mol C4H10/58.14 g C4H10) (10 mol H2O / 2 mol C4H10) ( 1.01 g H2O / 1 mol H2O ) = 0.15 g H2O

Calculate the mass of butane needed to produce 71.6 of carbon dioxide.
71.6 g CO2 (1 mol CO2/ 44.01 g CO2) ( 2 mol C4H10 / 8 mol CO2 ) (58.14 g C4H10 / 1 mol C4H10 ) = 23.65 g C4H10

4 0

3 years ago

Read 2 more answers

Using the Lewis dot structures of magnesium and oxygen, predict the ionic formula

Blizzard [7]

Answer:

(edit: nvm I figured it out, here is the answer)

Explanation: