Answer:
The answer to your question is 160 g of Fe₂O₃
Explanation:
Data
mass of Fe = 112 g
mass of CO = in excess
mass of Fe₂O₃ = ?
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
Process
1.- Calculate the molar mass of Fe₂O₃ and Fe
Molar mass Fe₂O₃ = (56 x 2) + (16 x 3) = 112 + 48 = 160 g
atomic mass of Fe = 56
2.- Use proportions to calculate the mass of Fe₂O₃ needed
160 g of Fe₂O₃ ------------------- 2(56) g of Fe
x g of Fe₂O₃ ------------------ 112 g of Fe
x = (112 x 160) / 2(56)
x = 17920/112
x = 160 g of Fe₂O₃
That's because the first and last carbon atoms cannot be branched to form an isomer . Therefore only the three middle carbon atoms can form isomers.
Answer:
21.5 g.
Explanation:
Hello!
In this case, since the reaction between the given compounds is:
We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:
Now, the moles of Li3P consumed by 15 g of Al2O3:
Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:
Therefore, the total mass of products is:
Which is not the same to the reactants (53 g) because there is an excess of Li₃P.
Best Regards!
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
The correct answer is option 2. A 0.8 M aqueous solution of NaCl has a higher boiling point and a lower freezing point than a 0.1 M aqueous solution of NaCl. This is explained by the colligative properties of solutions. For the two properties mentioned, the equation for the calculation of the depression and the elevation is expressed as: ΔT = -Km and <span>ΔT = Km, respectively. As we can see, concentration and the change in the property has a direct relationship.</span>