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Andreas93 [3]
2 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]2 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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The atomic weight of Ga is 69.72 amu. There are only two naturally occurring isotopes of gallium:
irga5000 [103]

Answer:

71 Ga has a naturally abundance of 36%

Explanation:

Step 1: Given data

Gallium has 2 naturally occurring isotopes: this means the abundance of the 2 isotopes together is 100 %. The atomic weight of Ga is 69.72 amu. This is the average of all the isotopes.

Since the average mass of 69.72 is closer to the mass of 69 Ga, this means 69 Ga will be more present than 71 Ga

Percentage 69 Ga> Percentage 71 Ga

<u>Step 2:</u> Calculate the abundance %

⇒Percentage of 71 Ga = X %

⇒Percentage of 69 Ga = 100 % - X %

The mass balance equation will be:

100*69.72 = x * 71 + (100 - x)*69

6972 = 71x + 6900 -69x

72 = 2x

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69 Ga has a naturally abundance of 64%

3 0
3 years ago
Which molecule is pentanoic acid?
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Answer:

A

Explanation:carboxyli

7 0
2 years ago
Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta
shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = \frac{5314.8}{2}\times 1=2657.4kJ

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

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3 years ago
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Why does Richard Rhodes think nuclear power is important moving forward?
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<em>Nuclear power releases less radiation into the environment than any other major energy source.</em>

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