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Andreas93 [3]
3 years ago
12

For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomas

sie Blue followed by the addition of 100 ml of 85% v/v phosphoric acid. This entire mixture is then diluted to 1 liter with water. What is the final concentration of phosphoric acid?
Chemistry
1 answer:
Flauer [41]3 years ago
4 0

Answer:

4.25% is the final concentration of phosphoric acid.

Explanation:

Initial concentration of phosphoric acid = C_1=85\%=0.85

Initial volume of phosphoric acid = V_1=50 mL

Final concentration of phosphoric acid = C_2=?

Final volume of phosphoric acid = V_2=1 L=1000 mL

( 1L = 1000 mL)

C_1V_1=C_2V_2

C_2=\frac{C_1times V_1}{V_2}

=\frac{0.85\times 50 mL}{1000 mL}=0.0425=4.25%

4.25% is the final concentration of phosphoric acid.

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How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C
rjkz [21]

20600Cal              

Explanation:

Given parameters:

Mass of water = 319.5g

Initial temperature = 35.7°C

Final temperature = 100°C

Unknown:

Calories needed to heat the water = ?

Solution:

The calories is the amount of heat added to the water. This can be determined using;

     H  =   m  c Ф

c  = specific heat capacity of water = 4.186J/g°C

   H is the amount of heat

    Ф is the change in temperature

    H = m c (Ф₂ - Ф₁)

    H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J

Now;

     1kilocalorie = 4184J

     

85996.56J to kCal; \frac{85996.56}{4184}   = 20.6kCal  = 20600Cal

               

learn more:

Specific heat brainly.com/question/3032746

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3 years ago
1. Long strands of DNA that contain lots of genes
sergeinik [125]

Answer: chromosome

Explanation: :)

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3 years ago
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In which volcanic eruption is pyroclastic materials released
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4 years ago
From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
Nina [5.8K]

<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

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dolphi86 [110]

Answer:

Natural resources

Explanation:

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