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2 years ago

Which would be the best material to use for making tea kettles?

Chemistry

1 answer:

Wewaii [24]2 years ago

6 0

It’s a subjective answer but glass is believed to be the most safe material to use for tea kettles

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1. Rust, or Fe2O3, is a Chemical Compound.

2. Carbon is a Chemical Element found as C on the periodic table.

3. Apple Juice is a Heterogeneous Mixture, as it is uniform throughout. If it wasn't uniform, there would be yucky lumps in your apple juice :)

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2 years ago

Example of transforming from electric energy to chemical energy<br> please!!!

Alik [6]

Answer:

Charging a chemical battery is a simple example of converting electricital energy to chemical energy.

Explanation:

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2 years ago

What happens when you increase the temperature of a reaction?

aalyn [17]

D. More collisions occur and the time required for the reaction decreases
This happens because according to collision theory, when energy (in this case, thermal energy) is applied to particles, they move/vibrate more quickly.

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3 years ago

Read 2 more answers

What is the heat in joules required to melt 25 grams of ice?

AfilCa [17]

Answer:

8,350 Joules

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

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3 years ago