Answer:
The rate at which sulfur dioxide is being produced is 0.90 kg/s.
Explanation:
Volume of oxygen gas consumed in second ,V= 994 L
Pressure of the gas = p
Temperature of the gas = T = 170°C= 170 + 273 K=443 K
Moles of oxygen gas consumed in a second = n
( ideapl gas equation)
n = 21.044 mole
Moles of dioxygen gas consumed per second = 21.044 mol
(Claus process)
According to reaction, 3 moles of dioxygen gives 2 moles of sulfur dioxide gas.Then 21.044 moles of dioxygen will give;
of sulfur dioxide
Mass of 14.029 moles of sulfur dioxide gas;
14.029 mol × 64 g/mol = 897.86 g
897.86 g = 0.89786 kg ≈ 0.90 kg
Mass of sulfur dioxide produced per second = 0.90 kg
The rate at which sulfur dioxide is being produced is 0.90 kg/s.
<u>Answer:</u> The number of moles of nitrogen gas is 0.030 moles
<u>Explanation:</u>
To calculate the number of moles of nitrogen, we use the equation given by ideal gas which follows:
where,
P = pressure of the nitrogen gas = 2.96 atm
V = Volume of the gas = 0.25 L
T = Temperature of the gas = 300 K
R = Gas constant = 
n = number of moles of nitrogen gas = ?
Putting values in above equation, we get:
Hence, the number of moles of nitrogen gas is 0.030 moles
Yes there is a great way to balance equation s
<h3>
Answer:</h3>
5.89 × 10^23 molecules of F₂
<h3>
Explanation:</h3>
The equation for the reaction between fluorine (F₂) and ammonia (NH₃) is given by;
5F₂ + 2NH₃ → N₂F₄ + 6 HF
We are given 66.6 g NH₃
We are required to determine the number of fluorine molecules
<h3>Step 1: Moles of Ammonia </h3>
Moles = Mass ÷ Molar mass
Molar mass of ammonia = 17.031 g/mol
Moles of NH₃ = 66.6 g ÷ 17.031 g/mol
= 3.911 moles
<h3>Step 2: Moles of Fluorine </h3>
From the equation 5 moles of Fluorine reacts with 2 moles of ammonia
Therefore,
Moles of fluorine = Moles of Ammonia × 5/2
= 3.911 moles × 5/2
= 9.778 moles
<h3>Step 3: Number of molecules of fluorine </h3>
We know that 1 mole of a compound contains number of molecules equivalent to the Avogadro's number, 6.022 × 10^23 molecules
Therefore;
1 mole of F₂ = 6.022 × 10^23 molecules
Thus,
9.778 moles of F₂ = 9.778 moles × 6.022 × 10^23 molecules/mole
= 5.89 × 10^23 molecules
Therefore, the number of fluorine molecules needed is 5.89 × 10^23 molecules
830 mL
The volume of an 2.3 m solution with 212 grams of calcium chloride (cacl2) dissolved is 830 mL.
The solution has a concentration of 2.3 mol/L.
<h3>a) Moles of CaCl2</h3>
Molar mass of CaCl2 = 110.98 g/mol
Moles of CaCl2 = 212 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2)
= 1.910 mol CaCl2
<h3>b) Volume of solution</h3>
V = 1.910 mol CaCl2 x (1 L solution/2.3 mol CaCl2) = 0.83 L solution
= 830 mL solution
<h3>How much CaCl2 is there in the solution by molarity?</h3>
- The number of moles is 0.125 x 2 = 0.25 mol since the molarity is 2.0M.
- To get the answer of 27.745 g, simply multiply this by the molar mass of calcium chloride, which is 110.98 g/mol.
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