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olchik [2.2K]
2 years ago
5

The stock concentration of dye is 3.4E-5M. The stock concentration of bleach is 0.36M. Assuming that the reaction goes to comple

tion (all the dye reacts), what is the remaining concentration of bleach, when 9.0 mL of dye reacts with 1.0 mL of bleach in 1:1 ratio? Give the answer with 2 SF
Chemistry
1 answer:
Aleks04 [339]2 years ago
5 0

Answer:

0.036 M

Explanation:

To do this, let's mark the dye as D and bleach as B.

We have the concentrations of both, and we already know that they react in a 1:1 mole ratio. The total volume of reaction is 9 + 1 = 10 mL or 0.010 L, and we hava both concentrations.

The problem already states that the dye reacts completely, so this is the limiting reagent, while bleach is the excess.

To know the remaining amount of bleach, we need to do this with the moles. First, let's calculate the initial moles of D and B:

moles D = 3.4x10⁻⁵ * 0.009 = 3.06x10⁻⁷ moles

moles B = 0.36 * 0.001 = 3.6x10⁻⁴ moles

Now that we have the moles, and that we know that all the dye reacts completely, let's see how many moles of bleach are left:

moles of B remaining = 3.6x10⁻⁴ - 3.06x10⁻⁷ = 3.597x10⁻⁴ moles

These are the moles presents of B after the reaction has been made. The concentration of the same will be:

[B] = 3.597x10⁻⁴ / 0.010

[B] = 0.0357

With 2 SF it would be:

[B] = 3.6x10⁻² M

You might be interested in
A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this p
xxTIMURxx [149]

Answer:

Limiting reagent = lead(II) nitrate

Theoretical yield = 3.75435 g

% yield = 65.26 %

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For potassium chloride :

Molarity = 1.20 M

Volume = 25.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of potassium chloride :

Moles=1.20 \times {25.0\times 10^{-3}}\ moles

<u>Moles of potassium chloride  = 0.03 moles</u>

For lead(II) nitrate :

Molarity = 0.900 M

Volume = 15.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of lead(II) nitrate :

Moles=0.900 \times {15.0\times 10^{-3}}\ moles

<u>Moles of lead(II) nitrate  = 0.0135 moles</u>

According to the given reaction:

2KCl_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbCl_2_{(s)}+2KNO_3_{(aq)}

2 moles of potassium chloride react with 1 mole of lead(II) nitrate

1 mole of potassium chloride react with 1/2 mole of lead(II) nitrate

0.03 moles potassium chloride react with 0.03/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.015 moles

<u>Limiting reagent is the one which is present in small amount. Thus, lead(II) nitrate is limiting reagent. (0.0135 < 0.015)</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) nitrate gives 1 mole of lead(II) chloride

0.0135 mole of lead(II) nitrate gives 0.0135 mole of lead(II) chloride

Molar mass of lead(II) chloride = 278.1 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.0135 × 278.1 g = 3.75435 g

<u>Theoretical yield = 3.75435 g</u>

Given experimental yield = 2.45 g

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (2.45/3.75435) × 100 = 65.26 %</u>

6 0
2 years ago
Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon
Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: Cu_2O\text{ and }CuO

  • <u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = \frac{2.26}{2.26}=1

For Oxygen  = \frac{4.55}{2.26}=2.01\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is CO_2

  • <u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = \frac{3.57}{3.57}=1

For Oxygen  = \frac{3.57}{3.57}=1

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is CO

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = \frac{1}{2}\times 1=\frac{1}{2}

Now, taking the ratio of carbon atoms in both the samples, we get:

C_1:C_2=\frac{1}{2}:1=1:2

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0
2 years ago
4. The stockroom contains 1.0 M NaAc (Sodium Acetate), 1.0 M HAc (acetic Acid), distilled water and strong acids and bases. You
Elina [12.6K]

Answer:

Concentrations of HAc and NaAc you need are 0.122M

Explanation:

pKa of acetic acid is 4.75, that means when amount of sodium acetate and acetic acid is the same, pH will be 4.75

Thus, you know [NaAc]i = [HAc]i

Now, using H-H equation, when pH = 3.75:

3.75 = 4.75 + log [NaAc] / [HAc]

0.1 = [NaAc] / [HAc]

10 [NaAc] = [HAc]

Thus, after the reaction  [HAc] must be ten times,  [NaAc].

Based in the reaction of NaAc with HCl

NaAc + HCl → HAc + NaCl

Moles of HCl added are:

1mL = 0.001L * (10mol /L) = 0.01 moles HCl.

That means moles of both compounds after the reaction are:

<em>[NaAc] = [NaAc]i - 0.01 mol </em>

[HAc] = [HAc]i + 0.01

Replacing these equations with the information you know:

[NaAc] = [NaAc]i - 0.01 mol

10[NaAc] = [NaAc]i + 0.01

Subtracting both equations:

9[NaAc] = 0.02mol

[NaAc] = 0.0022 moles.

Replacing in <em>[NaAc] = [NaAc]i - 0.01 mol </em>

0.0022mol = [NaAc]i - 0.01 mol

0.0122mol = [NaAc]i = [HAc]i

These moles in 100.00mL = 0.1000L:

[NaAc]i = [HAc]i = 0.0122mol / 0.100L =

0.122M

Thus, <em>concentrations of HAc and NaAc you need are 0.122M</em>

<em />

To create this buffer, you need to pipette 12.2mL of both 1.0M NaAc and 1.0M HAc and dilute this mixture to 100.0mL

4 0
2 years ago
Why is it advisable to wear long sleeves when students work in a chemistry lab? to provide warmth near the lab refrigerator to p
mote1985 [20]
It  is  advisable  to  wear  long  sleeve  when  when  a  student   is  working   in  a  chemistry  lab so   that   to  protect  arms   from  lab  chemicals. when   someone  enter  the  chemistry  lab  to  wort  or  to  study  should  be  well  prepared  with  appropriate  gears   and  security  measure   to  avoid   injury.
9 0
2 years ago
Read 2 more answers
Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −
Grace [21]

Answer : The value of K_{eq} is, 11.2

The value of \Delta G_{rxn} is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 25^oC=273+25=298K

K_{eq}  = equilibrium constant  = ?

Now put all the given values in the above formula, we get:

\Delta G^o=-RT\times \ln K_{eq}

-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}

K_{eq}=11.2

Thus, the value of K_{eq} is, 11.2

Now we have to calculate the \Delta G_{rxn}.

The formula used for \Delta G_{rxn} is:

The given reaction is:

A(aq)\rightleftharpoons B(aq)

\Delta G_{rxn}=\Delta G^o+RT\ln Q

\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}    ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction  = ?

\Delta G_^o =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})

\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol

Therefore, the value of \Delta G_{rxn} is -9.04 kJ/mol

3 0
2 years ago
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