<span> x + 2 = 6 x + 2 = 6 x+2=6x, plus, 2, equals, 6 has a variable in it.</span>
Answer:
75.15 g/mol
Explanation:
First, let us look at the equation of reaction;
From the balanced equation of reaction, 1 mole of NaOH is required to completely neutralize 1 mole of HAA.
Recall that: mole = molarity x volume.
Therefore, 27.50 mL, 0.120 M NaOH = 0.0275 x 0.120 = 0.0033 moles
0.0033 mole of NaOH will therefore requires 0.0033 moles of HAA for complete neutralization.
In order to find the molar mass of the unknown amino acid, recall that:
<em>mole = mass/molar mass</em>, hence, <em>molar mass = mass/mole</em>.
Therefore, molar mass of HAA = 0.248/0.0033 = 75.15 g/mol
Im confused here. Add the rest of the question for me to help you
Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
Mole-mole calculations are not the only type of calculations that can be performed using balanced chemical equations. Recall that the molar mass can be determined from a chemical formula and used as a conversion factor. We can add that conversion factor as another step in a calculation to make a mole-mass calculation, where we start with a given number of moles of a substance and calculate the mass of another substance involved in the chemical equation, or vice versa.
For example, suppose we have the balanced chemical equation
2 Al + 3 Cl 2 → 2 Alcoa
Suppose we know we have 123.2 g of Cl 2. How can we determine how many moles of Alcoa we will get when the reaction is complete? First and foremost, chemical equations are not balanced in terms of grams; they are balanced in terms of moles. So to use the balanced chemical equation to relate an amount of Cl 2 to an amount of Alcoa, we need to convert the given amount of Cl 2 into moles. We know how to do this by simply using the molar mass of Cl 2 as a conversion factor. The molar mass of Cl 2 (which we get from the atomic mass of Cl from the periodic table) is 70.90 g/mil. We must invert this fraction so that the units cancel properly:
