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3 years ago

You are working a crime scene and discover a weapon. How do you pick up the weapon so you do not disturb

Chemistry

2 answers:

11Alexandr11 [23.1K]3 years ago

3 0

Answer:

you pick it up with gloves and make minimal contact with it

Feliz [49]3 years ago

3 0

You would pick up the weapon with gloves from the bottom, this insures that you have minimal contact with it and not much DNA is wiped off.

You might be interested in

How many atoms of hydrogen are in 100 g of hydrogen peroxide (h2o2)?

Maurinko [17]

Hello friends..

find the MW of HP
calculate the # of mols per 100g of HP
take the # of mols times 6.023 x 10^23 times 2 (2 H per molecule)

Hope it helps you..

7 0

3 years ago

What is the mole fraction of O2O2 in a mixture of 15.1 gg of O2O2, 8.19 gg of N2N2, and 2.46 gg of H2H2

SVETLANKA909090 [29]

Answer:

Mole fraction O₂= 0.43

Explanation:

Mole fraction is the moles of gas/ total moles.

Let's determine the moles of each:

Moles O₂ → 15.1 g / 16 g/mol = 0.94

Moles N₂ → 8.19 g / 14 g/mol = 0.013

Moles H₂ → 2.46 / 2 g/mol = 1.23

Total moles = 2.183

Mole fraction O₂= 0.94 / 2.183 → 0.43

3 0

3 years ago

When 0.491 grams of a protein were dissolved in 44 mL of benzene at 24.4 degrees C, the osmotic pressure was found to be 50.9 to

Ira Lisetskai [31]

Answer:

4057.85 g/mol

Explanation:

Hello, the numerical procedure is shown in the attached file.

- In this case, since we don't have the density of the protein, we must assume that the volume of the solution is solely given by the benzene's volume, in order to obtain the moles of the solute (protein).

-Van't Hoff factor is assumed to be one.

Best regards.

4 0

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 $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

3 0

3 years ago

Based on your solubility rules which of the following compounds would form a precipitate water?

Novosadov [1.4K]

H2SO4+NaOH=Na2SO4+H2O

3 0

3 years ago