The nucleus of a comet is made up of rock, dust, and frozen gasses
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<h3>
Answer:</h3>
266.325 g
<h3>
Explanation:</h3>
We are given the balanced equation;
2NaOH + H₂SO₄ → H₂O + Na₂SO₄
We are required to determine the mass of Na₂SO₄ that will be formed.
<h3>Step 1: Determine the number of moles of NaOH</h3>
Moles = Mass ÷ molar mass
Molar mass of NaOH is 40.0 g/mol
Therefore;
Moles of NaOH = 150 g ÷ 40 g/mol
= 3.75 moles
<h3>Step 2: Determine the number of moles of sodium sulfate formed</h3>
- From the equation 2 moles of NaOH reacts with sulfuric acid to form 1 mole of sodium sulfate.
- Therefore; mole ratio of NaOH : Na₂SO₄ is 2 : 1
Thus, moles of Na₂SO₄ = Moles of NaOH ÷ 2
= 3.75 moles ÷ 2
= 1.875 moles
<h3>Step 3: Determine the mass of Na₂SO₄ produced.</h3>
we know that;
Mass = Moles × Molar mass
Molar mass of Na₂SO₄ is 142.04 g/mol
Therefore;
Mass of Na₂SO₄ = 1.875 moles × 142.04 g/mol
= 266.325 g
Thus, the mass of sodium sulfate formed 266.325 g
Answer:
The percentage yield is 78.2g
Explanation:
Given, mass of propane = 42.8 g , sufficient O2 percent yield = 61.0 % yield.
Reaction - C3H8(g)+5O2(g)------> 3CO2(g)+4H2O(g)
First we need to calculate the moles of propane
Moles of propane =
g.mol-1
= 0.971 moles
So, moles of CO2 from the moles of propane
1 mole of C3H8(g) = 3 moles of CO2(g)
So, 0.971 moles of C3H8(g) = ?
= 2.913 moles of CO2
So theoretical yield = 2.913 moles
44.0 g/mol
= 128.2 g
So, the actual mass of CO2 = percent yield
theoretical yield / 100 %
= 61.0 %
128.2 g / 100 %
= 78.2 g
the mass of CO2 that can be produced if the reaction of 42.8 g of propane and sufficient oxygen has a 61.0 % yield is 78.2 g
Answer:
The compound is N2O4
Explanation:
We have certain important pieces of information about the compound;
1) it is an oxide (a binary compound of nitrogen and oxygen)
2) there are no N-N bonds present
3) there are no O-O bonds present
Since it contains only nitrogen and oxygen then nitrogen accounts for 25.9% of the molecule by mass then oxygen should account for (100-25.9) = 74.1% oxygen
Relative atomic mass of oxygen = 16
Relative atomic mass of nitrogen = 14
We now deduce the empirical formula
Nitrogen. Oxygen
25.9/14. 74.1/16
1.85/1.85. 4.6/1.85 (divide through by the lowest ratio)
1 2
Empirical formula is NO2
To find the molecular formula
(NO2)n = 108
(14+2(16))n= 108
46n=108
n= 108/46
n= 2
Therefore molecular formula= N2O4