Answer:

Explanation:
Hello,
In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:

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Answer:
The final volume is 39.5 L = 0.0395 m³
Explanation:
Step 1: Data given
Initial temperature = 200 °C = 473 K
Volume = 0.0250 m³ = 25 L
Pressure = 1.50 *10^6 Pa
The pressure reduce to 0.950 *10^6 Pa
The temperature stays constant at 200 °C
Step 2: Calculate the volume
P1*V1 = P2*V2
⇒with P1 = the initial pressure = 1.50 * 10^6 Pa
⇒with V1 = the initial volume = 25 L
⇒with P2 = the final pressure = 0.950 * 10^6 Pa
⇒with V2 = the final volume = TO BE DETERMINED
1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2
V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)
V2 = 39.5 L = 0.0395 m³
The final volume is 39.5 L = 0.0395 m³
Answer:
(II) only correctly rank the bonds in terms of increasing polarity.
Explanation:
Bond polarity is proportional to difference in electronegativity between bonded atoms.
Atoms Electronegativity Bond Electronegativity difference
Cl 3.0 Cl-F 1.0
Br 2.8 Br-Cl 0.2
F 4.0 Cl-Cl 0
H 2.1 H-C 0.4
C 2.5 H-N 0.9
N 3.0 H-O 1.4
O 3.5 Br-F 1.2
I 2.7 I-F 1.3
Si 1.9 Cl-F 1.0
P 2.2 Si-Cl 1.1
Si-P 0.3
Si-C 0.6
Si-F 2.1
So, clearly, order of increasing polarity : O-H > N-H > C-H
So, (II) only correctly rank the bonds in terms of increasing polarity
Answer:- 64015 J
Solution: There is 4250 mL of water in the calorimeter at 22.55 degree C.
density of water is 1 g per mL.
So, the mass of water =
= 4250 g
Final temperature of water after adding the hot copper bar to it is 26.15 degree C.
So,
for water = 26.15 - 22.55 = 3.60 degree C
Specific heat for water is 4.184 
The heat gained by water is calculated by using the formula:

where, q is the heat energy, m is mass and c is specific heat.
Let's plug in the values in the formula and do the calculations:

q = 64015 J
So, 64015 J of heat is gained by the water.
To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:
N₂ + O₂ → 2 NO
Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.
1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present
Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>