Answer: [B]: 12.0 .
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Note: The pH scale ranges from 1 to 14 .
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Answer:
V NH3 = 304.334 L
Explanation:
- 1 mol ≡ 6.02 E23 molecules
⇒ moles NH3 = (7.50 E24 molecules)×(mol/6.022 E23 molecules)
⇒ mole NH3 = 12.454 mol
assuming ideal gas:
STP:
∴ T = 25°C ≅ 298 K
∴ P = 1 atm
⇒ V NH3 = RTn/P
⇒ V NH3 = ((0.082 atm.L/K.mol)×(298 K)×(12.454 mol))/(1 atm)
⇒ V NH3 = 304.334 L
Answer:
1.21 g of Tris
Explanation:
Our solution if made of a solute named Tris
Molecular weight of Tris is 121 g/mol
[Tris] = 100 mM
This is the concentration of solution:
(100 mmoles of Tris in 1 mL of solution) . 1000
Notice that mM = M . 1000 We convert from mM to M
100 mM . 1 M / 1000 mM = 0.1 M
M = molarity (moles of solute in 1 L of solution, or mmoles of solute in 1 mL of solution). Let's determine the mmoles of Tris
0.1 M = mmoles of Tris / 100 mL
mmoles of Tris = 100 mL . 0.1 M → 10 mmoles
We convert mmoles to moles → 10 mmol . 1mol / 1000mmoles = 0.010 mol
And now we determine the mass of solute, by molecular weight
0.010 mol . 121 g /mol = 1.21 g
Answer:
1) The concentration of HCl = 0.1 M.
2) The table that can be used to organize the information correctly is C.
Explanation:
<u><em>1) The concentration of HCl:</em></u>
- We know that the no. of millimoles of the acid is equal to the no. of millimoles of the base at the neutralization point.
which means that: <em>(MV)HCl = (MV)NaOH,</em>
M of HCl = ??? M, V of HCl = 25.0 mL.
M of NaOH = 0.2 M, V of NaOH = 12.5 mL.
∴ M of HCl = (MV)NaOH/V of HCl = (0.2 M)(12.5 mL)/(25.0 mL) = 0.1 M.
<em>2) The table that can be used to organize the information correctly is C</em>
<em></em>
<em>Table A and B are the same and reported volume of HCl and NaOH is wrong.</em>
<em>Table C is right, contain the correct volumes and concentration of NaOH and missed the concentration of HCl which is 0.1 M.</em>
<em>Table D reported the volume and the concentration of HCl wrongly and also the concentration of NaOH. The data reported of HCl and NaOH is reversed.</em>
All you need to do the cross the charges and put it as suffixes.
D+2 + E-1 ----> DE2
answer is C
