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Sergio039 [100]
3 years ago
12

What is the pH of a solution with (H+) = .01 M? Show your work.​

Chemistry
1 answer:
slamgirl [31]3 years ago
6 0

Answer:

pH=2

Explanation: pH= −lg([H+]) = -lg([0,01]) = 2

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Q: A 25.5 mL aliquot of HCl (aq) of unknown concentration was titrated with 0.113 M NaOH (aq). It took 51.2 mL of the base to re
blondinia [14]
M ( HCl ) = ?

V ( HCl ) = 25.5 mL in liters : 25.5 / 1000 => 0.0255 L

M ( NaOH ) = 0.113 M

V ( NaOH ) = 51.2 mL / 1000 => 0.0512 L

number of moles NaOH:

n = M x V

n = 0.113 x <span> 0.0512 => 0.0057856 moles of NaOH

mole ratio:

</span><span>HCl + NaOH = NaCl + H2O
</span><span>
1 mole HCl -------------- 1 mole NaOH
( moles HCl ) ----------- </span><span> 0.0057856 moles NaOH
</span>
(moles HCl ) = <span> 0.0057856 x 1 / 1
</span>
= <span> 0.0057856 moles of HCl
</span>
M ( HCl ) = n / V

M =  0.0057856 / <span>0.0255
</span>
= 0.227 M

Answer A

hope this helps!

4 0
3 years ago
Read 2 more answers
I have a 4 amino acid peptide of unknown sequence with a curious titration profile around pH = 6. If Iknow that the peptide N-te
netineya [11]

Answer:

c. Histidine

Explanation:

Histidine is a compound that is normally used for the generation of protein. Three amino acids commonly have basic side chain when the pH is neutral. The conjugate acid in histidine has a pKa of approximately 6. Based on the description of the experimental analysis provided in the statement, the right option is option c.

4 0
4 years ago
The Kp for the reaction below is 1.49 × 108 at 100.0°C:
Anna007 [38]
Kp= (COCl2)/[(CO)(Cl2)]= 1.49 x 10^8

1.49 x 10^8= (COCl2/((2.22x10-4)(2.22x10-4))

COCl2= 1.49x10^8 x ((2.22x10-4)(2.22x10-4))= 7.34 atm
5 0
3 years ago
What is the mass in grams of a single formula unit of silver chloride, AgCI? A) 4.21 x 1021 g B) 8.61 x 10258 C) 1.66 x 10-248 D
bekas [8.4K]

Answer: D) 2.38*10^-^2^2 g

Explanation: The question asks to convert formula unit to grams. It is a unit conversion problem.

1 mole equals to Avogadro number of formula units. So, to convert the given number of formula units to moles, we need to divide by the Avogadro number. After this, we do moles to grams conversion and for this the moles are multiplied by the molar mass of the compound. Molar mass of AgCl is 143.32 gram per mol.

1FormulaUnitAgCl(\frac{1mol}{6.022*10^2^3formulaUnits})(\frac{143.32g}{1mol})

= 2.38*10^-^2^2 g

So, the correct option is D) 2.38*10^-^2^2 g

7 0
3 years ago
Read 2 more answers
Question<br> What is the molarity of a 400 mL solution containing 0.60 moles of NaCl?
Mumz [18]

Answer:

0.24 M

Explanation:

Molarity = Moles solute / Liters solution

Step 1: Identify variables

400 mL = Liters solution

0.60 moles = Moles solute

Step 2: Identify conversions

1 L = 1000 mL

Step 3: Convert mL to L

400mL(1 L/1000mL) = 0.4 L

Step 4: Find molarity

M = (0.4 L)(0.60 mol) = 0.24 M

6 0
3 years ago
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