Answer:
972.3 Torr
Explanation:
P2=P1V1/V2
You can check this by knowing that P and V at constant T have an an inverse relationship. Hence, this is correct.
Answer:
The answer to your question is
1.-Fe₂O₃
2.- 280 g
3.- 330 g
Explanation:
Data
mass of CO = 224 g
mass of Fe₂O₃ = 400 g
mass of Fe = ?
mass of CO₂
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
1.- Calculate the molar mass of Fe₂O₃ and CO
Fe₂O₃ = (56 x 2) + (16 x 3) = 160 g
CO = 12 + 16 = 28 g
2.- Calculate the proportions
theoretical proportion Fe₂O₃ /3CO = 160/84 = 1.90
experimental proportion Fe₂O₃ / CO = 400/224 = 1.78
As the experimental proportion is lower than the theoretical, we conclude that the Fe₂O₃ is the limiting reactant.
3.- 160 g of Fe₂O₃ --------------- 2(56) g of Fe
400 g of Fe₂O₃ --------------- x
x = (400 x 112) / 160
x = 280 g of Fe
4.- 160 g of Fe₂O₃ --------------- 3(44) g of CO₂
400 g of Fe₂O₃ -------------- x
x = (400 x 132)/160
x = 330 gr
C. Magma from venus mantle erupted as lava.
Explanation:
A volcano is a land form which results from the eruption of molten rocks (lava) on the surface. Volcanic rocks are a special type of igneous rock that forms when molten rock cools and solidifies on the surface.
For a planet like Venus which is presently not active and little to no movement occurs within the plates, the volcanisim must have occurred when the planet was relatively young and it must have been millions of years ago.
It is widely believed that Venus was geologically active in times past. Mantle generated lava must have solidified on the surface in times past to have formed the volcano.
Evaluating other options:
Impact of space objects on Venus would lead to the formation of a crater which is a depression on the surface. The rock would be mostly metamorphic.
If water was ever present in Venus, they would have produced sedimentary rocks instead. The erosive power of water is not high enough to cut through the crust. Also, water would not aid the formation of volcanoes.
Heat is not enough to from volcanoes. Other factors are also in play.
It’s 5% because without the 5% you wouldn’t make it to 100 equally
The percentage of the sulfur (S) in the compound CuSO₄ is 20.1 %.
<h3>What is the mass percentage?</h3>
The percentage of an element in a compound can be determined as the number of parts by mass of that element present in 100 parts by mass of the given compound.
First, calculate the molecular mass of the given compound by the addition of the atomic masses of all the present elements in the molecular formula. Then, the percentage of the elements can be determined by dividing the total mass of the element by the molar mass of the compound multiplied by 100.
Given, the atomic mass of copper, sulfur, and oxygen is 63.55 g, 32.07 g, and 16.0g respectively.
The molecular mass of CuSO₄ = 63.55 + 32.07 + 4(16.0) = 159.62 g
The mass percentage of the sulphur = (32.07/159.62) × 100 = 20.1 %
Therefore, the mass percentage of the sulfur is equal to 20.1 %.
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