Answer:
<h3>A. Epimers </h3>
Explanation:
<h2>Hope it help ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️</h2>
According to one acid-base theory, a water molecule acts as an acid when the water molecule (3) donates an H+.
Answer:
Explanation:
The air 9% mole% methane have an average molecular weight of:
9%×16,04g/mol + 91%×29g/mol = 27,8g/mol
And a flow of 700000g/h÷27,8g/mol = 25180 mol/h
In the reactor where methane solution and air are mixed:
In = Out
Air balance:
91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)
Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>
<em></em>
The air in the product gas is
95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>
43058 mol air×29g/mol <em>1249 kg air</em>
Percent of oxygen is: 
=<em>0,231 kg O₂/ kg air</em>
<em></em>
I hope it helps!
Answer:
6.4 g BaSO₄
Explanation:
You have been given the molarity and the volume of the solution. To find the mass of the solution, you need to (1) find the moles BaSO₄ (via the molarity ratio) and then (2) convert moles BaSO₄ to grams BaSO₄ (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given values.
Molarity (mol/L) = moles / volume (L)
(Step 1)
55 mL / 1,000 = 0.055 L
Molarity = moles / volume <----- Molarity ratio
0.5 (mol/L) = moles / 0.055 L <----- Insert values
0.0275 = moles <----- Multiply both sides by 0.055
(Step 2)
Molar Mass (BaSO₄): 137.33 g/mol + 32.065 g/mol + 4(15.998 g/mol)
Molar Mass (BaSO₄): 233.387 g/mol
0.0275 moles BaSO₄ 233.387 g
--------------------------------- x ------------------- = 6.4 g BaSO₄
1 mole
