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pickupchik [31]
2 years ago
13

Can someone help with this problem

Chemistry
1 answer:
gavmur [86]2 years ago
3 0
1. Fe2O3
2. Molar mass of Fe2O3 - (55.8x2)+(16x3) = 159.6
Moles =mass/Molar mass
Moles=9.2/159.6=0.058mol
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Gold has a density of 19.32 is this a chemical change or physical change
forsale [732]

Answer:

chemical change it is melted down

7 0
3 years ago
the elements iron (Fe), copper (Cu), and mercury (Hg) are classified as metals. which physical property of metals do they all sh
koban [17]

Answer:

They are heavy metals.

Explanation:

Heavy metals are generally defined as metals with relatively high densities, atomic weights, or atomic numbers.

6 0
3 years ago
How much energy is evolved during the formation of 197 g of Fe, according to the reaction below?
hoa [83]

Answer:

ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'

Explanation:

Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj

197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)

From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...

3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).

______

NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g.  Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.  

8 0
2 years ago
if the relationship between tempurature and volume of a gas at a constant pressure is linear, explain why the volume of a gas do
Lilit [14]
The gas has already created enough pressure to become a gas, which is the most expandable it could turn in to. Thats what i think, hope it helps.
4 0
2 years ago
How much energy (kJ) is required to change 0.18 mole of ice (s) at 0 C to water (l) at 0 C?
Dmitriy789 [7]

Answer:25,06 kJ of energy must be added to a 75 g block of ice.

ΔHfusion(H₂O) = 6,01 kJ/mol.

T(H₂O) = 0°C.

m(H₂O) = 75 g.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 75 g ÷ 18 g/mol.

n(H₂O) = 4,17 mol.

Q = ΔHfusion(H₂O) · n(H₂O)

Q = 6,01 kJ/mol · 4,17 mol

Q = 25,06 kJ.

Explanation:

6 0
2 years ago
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