Which of the following procedures demonstrates repetition? A. Jeremiah measures the mass of five different strips of metal. He p
2 answers:
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Answer:
It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.
Explanation:
The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C or 273.15 °K are used and are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P*V = n*R*T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.
So, in this case:
- P= 1 atm
- V= 855 L
- n= ?
- R= 0.082
- T= 273.15 K
Replacing:
1 atm* 855 L= n* 0.082 * 273.15 K
Solving:
n= 38.17 moles
Being the molar mass of nitrogen N2 equal to 28 g / mol, you can apply the following rule of three: if there are 28 grams in 1 mole, how much mass is there in 38.17 moles?
mass= 1,068.76 grams
<u><em> It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.</em></u>
The empirical formula of the initial zinc oxide is ZnO.
<h3>What is Empirical Formula?</h3>
The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.
It is the lowest whole number ratio of the element in the compound.
<h3>How to find out the empirical formula?</h3>- Find out the given masses and molar masses of the elements
The molar mass of Zn = 65 gmol⁻¹
Given the mass of Zn = 2.156 g
The molar mass of Oxygen = 16 gmol⁻¹
The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal
= (2.684 - 2.156) g
= 0.528 g
- Find the number of moles of the elements in the compound
The number of moles is given by
where m = given mass and
M = Molar mass
Number of moles of Zinc = = 0.033 moles
Number of moles of Oxygen = = 0.033 moles
- Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.
Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.
Therefore, the empirical formula of zinc oxide is ZnO.
Learn more about the empirical formula:
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