Answer:
but iron is 2 and 3 by its valance how could it be possible
The density of the metal with a mass of 1.71g that was dropped into a graduated cylinder containing 17.00 mL of water is 1.005g/mL.
<h3>How to calculate density?</h3>
The density of a substance can be calculated by dividing the mass of the substance by its volume. That is;
Density = mass ÷ volume
According to this question, a piece of metal with a mass of 17.1 g was dropped into a graduated cylinder containing 17.00 mL of water. The density can be calculated as follows:
Density = 17.1g ÷ 17.00mL
Density = 1.005g/mL
Therefore, the density of the metal with a mass of 1.71g that was dropped into a graduated cylinder containing 17.00 mL of water is 1.005g/mL.
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Answer:
i believe the answer is a
Explanation:
Answer:
The answer to the question is
The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits
Explanation:
To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure
At the first trial the mixture contains
3.6 atm CO
1.2 atm H₂O (g)
Total pressure = 3.6+1.2= 4.8 atm
which gives
3.36 atm CO
0.96 atm H₂O (g)
0.24 atm H₂ (g)
That is
CO+H₂O→CO(g)+H₂ (g)
therefore the mixture contained
0.24 atm CO₂ and the total pressure =
3.36+0.96+0.24+0.24 = 4.8 atm
when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂
At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857
adding 1.8 atm CO gives 4.46 atm hence we have
(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857
which gives x = 0.031 atm or x = -0.6183 atm
Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm
