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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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Answer: Ok I think The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Carbon, or 12.0107 grams. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between moles Carbon and gram. 12 grams

This means that the atomic mass or atomic weight (12 grams) of carbon is equal to exactly 1 mole of carbon.

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Explanation:

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Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

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pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

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The number of moles of the gas when the reacting mass is 3.0 g;

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PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

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3 years ago
A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe
nikitadnepr [17]

Answer:

\large \boxed{34.2\, ^{\circ}\text{C}}

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:  

For the metal:

m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}

For the water:

m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}

\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}

\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}

\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}

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Results in the complete loss of an organism's home.
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Answer:

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Because the organisms habitat has been destroyed, (by many reasons such as Deforestation and Climate Change), the capacity to hold the current population of the plant or animals is declined which can eventually lead to the scenario of EXTINCTION

Explanation:

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