Calculate the Kelvin temperatures of the water and record your answers in the data table.

What is the matter of science

weqwewe [10]

The finagling in the hole

3 0

2 years ago

In the Polyacrylamide and Agarose Gels lab, when working with your protein or DNA sample you must wear:

blagie [28]

It’s a. All of these, you’re welcome

6 0

3 years ago

Read 2 more answers

How is primary selection different from secondary succession?

PIT_PIT [208]

Answer:

Explanation:

Primary and secondary succession occur after both human and natural events that cause drastic change in the makeup of an area. Primary succession occurs in areas where there is no soil and secondary succession occurs in areas where there is soil.

In primary succession, newly exposed or newly formed rock is colonized by living things for the first time. In secondary succession, an area previously occupied by living things is disturbed—disrupted—then recolonized following the disturbance.

Hope this helped :)

-<em>Akito</em>

8 0

2 years ago

The law of constant composition states: All atoms of a given element have a constant composition and are different than atoms of

Airida [17]

Answer:

C

Explanation:

The law proves C. For examples no matter how water you have it will always have a 1:2 ratio of oxygen to hydrogen. :)

7 0

2 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago