Answer:
P₂ = 394.4 KPa
Explanation:
Given data:
Volume of gas = 152 dm³
Pressure of gas = 98.6 KPa
Final pressure = ?
Final volume = quartered = 1/4×152 = 38 dm³
Solution:
P₁V₁ = P₂V₂
P₂ = P₁V₁/V₂
P₂ = 98.6 KPa . 152 dm³ / 38 dm³
P₂ = 14987.2 KPa. dm³ / 38 dm³
P₂ = 394.4 KPa
Answer:
arteries carry blood from the heart to the lungs and veins carry blood from the lungs to the heart
Explanation:
Arteries carry oxygenated blood from the heart, while veins carry oxygen-depleted blood back to the heart. ... The pulmonary veins transport oxygenated blood back to the heart from the lungs, while the pulmonary arteries move deoxygenated blood from the heart to the lungs.)
Answer:
Answer:
The mole ratio of C₄H₁₀ and CO₂ is 2 : 8, which simplifies to 1 : 4.
Explanation:
The mole ratio is the relative proportion of the moles of products or reactants that participate in the reaction according to the chemical equation.
The chemical equation given is:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Once you check that the equation is balanced, you can set the mole ratios for all the reactants and products. The coefficients used in front of each reactant and product, in the balanced chemical equation, tells the mole ratios.
In this case, they are: 2 mol C₄H₁₀ : 13 mol O₂ : 8 mol CO₂ : 10 mol H₂O
Since you are asked about the mole ratio of C₄H₁₀ and CO₂ it is:
2 mol C₄H₁₀ : 8 mol CO₂ , which dividing by 2, simplifies to
1 mol C₄H₁₀ : 4 mol CO₂, or
1 : 2.
Explanation:
Answer:
Today, the average total production cost of “standard modulus” carbon fiber is in the range of $7-9 per pound.
Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.
