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olchik [2.2K]
1 year 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]1 year 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

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Octane is a liquid component of gasoline. Given the following vapor pressures of octane at various temperatures, estimate the bo
Hitman42 [59]

Answer:

110.8 ºC

Explanation:

To solve this problem we will make use of the Clausius-Clayperon equation:

lnP = - ΔHºvap/RT + C

where P is the pressure, ΔHºvap is the enthalpy of vaporization, R is the gas constant, T is the temperature, and C is a constant of integration.

Now this equation has a form y = mx + b where

y = lnP

x = 1/T

m = -ΔHºvap/R

Now we have to assume that ΔHºvap remains constant which is a good asumption given the narrow range of temperatures in the data ( 104-125) ºC

Thus what we have to do is find the equation of the best fit for this data using a  software as excel or your calculator.

T ( K)               1/T                  ln P

377               0.002653       5.9915

384              0.002604       6.2115

390              0.002564       6.3969

395              0.002532       6.5511

398              0.002513        6.6333

The best line has a fit:

y = -4609.5 x  + 18.218

with R² = 0.9998

Now that we have the equation of the line, we simply will substitute for a pressure of 496 mm in Leadville.

ln(496) = -4609.5(1/Tb) + 18.218

6.2066 = -4609.5(1/Tb) +18.218

⇒ 1/Tb = (18.218 - 6.2066)/4609.5 = 0.00261

Tb = 383.76 K  = (383.76 -273)K = 110.8 ºC

Notice we have touse up to 4 decimal places since rounding could lead to an erroneous answer ( i.e boiling temperature greater than 111, an impossibility given the data in the question). This is as a result of the value 496 mmHg so close to 500 mm Hg.

Perhaps that is the reason the question was flagged.

7 0
2 years ago
(f) what is the observed rotation of 100 ml of a solution that contains 0.01 mole of d and 0.005 mole of l? (assume a 1-dm path
never [62]
<span>Answer: .01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L. so .01/.015 to .005/.015 ~ 67% D to 33% L. And thus, the enantiomer excess will be 34%.</span>
4 0
2 years ago
Read 2 more answers
An ice cube measuring 5.80 cm by 5.80 cm by 5.80 cm has a density of 0.917 g/mL What is the mass?
OLga [1]

Answer: 178.9 g

Explanation:

Density = \frac{mass}{volume in mL}

find volume of the cube: (5.80 cm) (5.80 cm) (5.80cm) = 195.112 cm³

1.0 cm³ = 1.0 mL

so 195.112 cm³ = 195.112 mL

plug value into density equation:

0.917 g/mL = (mass) / (195.112 mL)

and solve for mass!

3 0
1 year ago
How can substance A and B be compared? Choose all that Apply.
77julia77 [94]

I would say the 1.

2. and 4.

Hope this helps!

7 0
2 years ago
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Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l
Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

We are given:

\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol

Putting values in above equation, we get:

-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]

\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0
2 years ago
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