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3 years ago

CHEMISTRY PLEASE HELP!!!???

Chemistry

1 answer:

Montano1993 [528]3 years ago

5 0

1. Solids

- definite volume & shape

- little energy

-vibrate in place

- very incompressible

2. Liquids

- held together yet can still flow

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Can we use electrical equipment near water? Explain Why?

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Answer:

You may, but it is too risky.

Even though you are being cautious around using electric equipment around water, you'll never know what can happen. You might accidentally drop that piece of electrical equipment you are using into the water. Water can be splashed around by someone or something without you noticing it and it may affect the object you are using. Sometimes, if water comes in contact with an electrical object, it may cause you electric shocks or the equipment you are using has a chance of exploding and may hurt you. You can guarantee that waterproof electrical equipment is safe to use, but it is better not to risk it too much.

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3 years ago

3) In the Hydrolysis of Disaccharides and Polysaccharides portion of the lab, starch should give one positive iodine test and on

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Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.

Explanation:

Starch is an example of polysaccharide and since the standard test for it is potassium iodide solution, it gives a positive test.

Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .

Therefore, in the hydrolysis; starch should give a positve test, while Diasaccharides should give negative rest.

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3 years ago

Convert 15 meters to kilometers

Gelneren [198K]

Answer:

the answer is 0.015

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Read 2 more answers

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

How do the particles that are dispersed in a colloid differ from the particles suspended in a suspension?

Fed [463]

Answer:B, the colloid particles are larger.

PLEASE GIVE BRAINLIEST AND WATCH OUT BEHIND YOU

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3 years ago