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ludmilkaskok [199]
3 years ago
9

A sample of bleach was analyzed as in this procedure. The only procedural difference is that the student weighed out the bleach

solution used instead of pipetting a certain volume of bleach. The student weighed out 0.634 g of commercial bleach solution. It was found that it required 13.24 mL of 0.0732 M sodium thiosulfate solution to react with the iodine produced. What is the percentage of sodium hypochlorite in this bleach sample
Chemistry
1 answer:
Bogdan [553]3 years ago
6 0

Answer:

% = 5.69%

Explanation:

To do this, we need to write the equations taking place here. First, this is a REDOX reaction where the hypoclorite and thiosulfate solution reacts. The balanced equations are:

ClO⁻ + 2I⁻ + 2H⁺ -------> Cl⁻ +  I₂ + H₂O

I₂ + 2S₂O₃²⁻ -----------> 2I⁻ + S₄O₆²⁻

We already have the required volume and concentration of the thiosulfate solution, so we can calculate the moles of thiosulfate. With this moles, we can calculate the moles of hypochlorite, then the mass and finally the %.

The moles of thiosulfate would be:

moles S₂O₃²⁻ = V * M

moles S₂O₃²⁻ = 0.01324 * 0.0732 = 9.69x10⁻⁴ moles

Now according to the above reactions, we can see that

moles I₂ = moles ClO⁻

and

moles I₂ / moles S₂O₃²⁻ = 1/2

Therefore, let's calculate the moles of ClO⁻:

moles ClO⁻ = 9.69x10⁻⁴ / 2 = 4.845x10⁻⁴ moles

Now, we can calculate the mass of these moles, using the molar mass of sodium hypochlorite which is 74.44 g/mol:

m = 74.44 * 4.845x10⁻⁴

m = 0.036 g

Finally the % of this, in the bleach sample would be:

% = 0.036 / 0.634 * 100

<h2>% = 5.69%</h2>
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lora16 [44]
Find the molar mass of CH3 and divide that by 45.0. That should give a whole number and then mult that whole number by CH3 to find molecular formula to get like ( incorrect ex: C2H6) which is mult by 2
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2 years ago
An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterize
oee [108]

Answer:

(a) r = 6.26 * 10⁻⁷cm

(b) r₂ = 6.05 * 10⁻⁷cm

Explanation:

Using the sedimentation coefficient formula;

s =  M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle

s = M ( 1 - Vρ) / N*6πnr

making r sbjct of formula, r =  M (1 - Vρ) / N*6πnrs

Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s

r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)

r = 6.26 * 10⁻⁷cm

b. Using the formula r₂/r₁ = s₁/s₂

s₂ = 0.035 + 1s₁ = 1.035s₁

making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

r₂ = 6.3 * 10⁻⁷cm / 1.035

r₂ = 6.05 * 10⁻⁷cm

8 0
3 years ago
Tin has ten stable isotopes. The heaviest, 124Sn, makes up 5.80% of naturally occuring tin atoms. How many atoms of 124Sn are pr
matrenka [14]

The average atomic mass of Sn is 118.71 g/mol

the percentage of heaviest Sn is 5.80%

the given mass of Sn is 82g

The total  moles of Sn will be = mass / atomic mass = 82/118.71=0.691

Total atoms of Sn in 82g = 6.023X10^{23}X0.691=4.16X10^{23}

the percentage of heaviest Sn is 5.80%

So the total atoms of Sn^{124} = 5.80% X 4.16X10^{23}

Total atoms of Sn^{124}=2.41X10^{22} atoms

the mass of Sn^{124} will be = \frac{2.41X10^{22}X124}{6.023X10^{23}}=4.96g

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Yield for this reaction?<br> Reaction: N2(g) + 3 H2(g) → 2 NH3 (g)
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Answer:

Yes, yield.

Explanation:

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First, find limiting reactant:  

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Moles N2 = 9.84 g N2 x 1 mole/28 g = 0.351 moles N2

The mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)

Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3

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klasskru [66]

Your answer would be 0.024951344877489 but rounding it would be 0.025 moles

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