Answer:
The answer is "
"
Explanation:
Please find the complete question in the attached file.
Equation:
at 
at equilibrium 
![p= 0.47 \ \ atm\\\\SO_2=3.3-0.47 = 2.83 \ \ atm\\\\O_2= 0.74 -\frac{0.47}{2}=0.74-0.235=0.555 \ atm\\\\K_P=\frac{[PSO_3]^2}{[PSO_2]^2[PO_2]}\\\\](https://tex.z-dn.net/?f=p%3D%200.47%20%5C%20%5C%20atm%5C%5C%5C%5CSO_2%3D3.3-0.47%20%3D%202.83%20%5C%20%5C%20atm%5C%5C%5C%5CO_2%3D%200.74%20-%5Cfrac%7B0.47%7D%7B2%7D%3D0.74-0.235%3D0.555%20%5C%20atm%5C%5C%5C%5CK_P%3D%5Cfrac%7B%5BPSO_3%5D%5E2%7D%7B%5BPSO_2%5D%5E2%5BPO_2%5D%7D%5C%5C%5C%5C)
Answer:
19.8 kg of C₂H₂ is needed
Explanation:
We solve this by a rule of three:
If 1251 kJ of heat are relased in the combustion of 1 mol of acetylene
95.5×10⁴ kJ of heat may be released by the combustion of
(95.5×10⁴ kJ . 1) /1251kJ = 763.4 moles of C₂H₂
Let's convert the moles to mass → 763.4 mol . 26 g/1 mol = 19848 g
If we convert the mass from g to kg → 19848 g . 1kg / 1000g = 19.8 kg
Simple cations are formally called by their element names with a suffixed Roman numeral in parentheses to indicate its charge. A simple anion has a name that is the original elemental name with the final syllable changed to -ide.
<h3>
Answer:</h3>
82.11%
<h3>
Explanation:</h3>
We are given;
- Theoretical mass of the product is 137.5 g
- Actual mass of the product is 112.9 g
We are supposed to calculate the percentage yield
- We need to know how percentage yield is calculated;
- To calculate the percentage yield we get the ratio of the actual mass to theoretical mass and express it as a percentage.
Thus;
% yield = (Actual mass ÷ Experimental mass) × 100%
= (112.9 g ÷ 137.5 g) × 100%
= 82.11%
Therefore, the percentage yield of the product is 82.11 %
This may help you
First write and balance the equation, being:
CaCO3 - CaO + CO2
Then, using the periodic table, find the molecular masses of CaCO3 and of CaO, finding their ratio. That will be 100g:56g or 0.1kg:0.056kg. Since you have 4.7kg of CaCO3, it corresponds to Xkg of CaO. Making x the subject, it should be X= 4.7*0.056/100=0,002632
