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Marat540 [252]
3 years ago
10

How many ml of 0.01-m naoh will be required to titrate 0.061-g of khp to it's endpoint?

Chemistry
1 answer:
hjlf3 years ago
6 0
The volume of NaOH needed for the titration is calculated by equating the number of moles of KHP and NAOH.

(1) number of moles of KHP:
         n = mass of KHP / molar mass of KHP
Substituting,
         n = 0.061 g / (204.22 g/mol) = 2.987 x 10⁻⁴ moles KHP

(2) number of moles of NaOH
         n = (volume NaOH in L)(molarity of NaOH)
          n = V x 0.01

Equating,
       0.01V = 2.987 x 10⁻⁴

The value of V from the equation is 0.0299L of NaOH. 

Converting the value to mL will give us the answer of 29.87 mL. 
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2.5 × 10² ppm

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Step 1: Given data

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  • Mass of the sample: 200. g

Step 2: Convert 0.050 g to μg

We will use the conversion factor 1 g = 10⁶ μg.

0.050 g × 10⁶ μg/1 g = 5.0 × 10⁴ μg

Step 3: Calculate the concentration of NaCl in ppm

The concentration of NaCl in ppm is equal to the micrograms of NaCl per gram of the sample.

5.0 × 10⁴ μg NaCl/200. g = 2.5 × 10² ppm

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In this experiment, the experimental group is ice, the independent variable is sunlight while on the other hand, the dependent variable is block of ice.

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How many moles of chlorine (Cl) atoms are in a sample of 1.72 × 1022 atoms? 0.0286 mol Cl 35.0 mol Cl 1.03 × 1023 mol Cl 1.04 ×
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Given information : Sample of 1.72\times 10^{22} atoms

We need to find the moles of Chlorine in the given sample.

We can say that we need to find moles from the given atoms.

Relation between mole and atom is given by : 1 mole = 6.022\times 10^{23} atoms

Where 6.022\times 10^{23} is Avogadro number.

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