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3 years ago

27. Interpreting Concepts One way to make lemonade is

Chemistry

1 answer:

Fynjy0 [20]3 years ago

8 0

Explanation:

compound is already together use your brain man you take multiple things. to make lemonade aka a mixture of powder and water cmon man u gotta think about it logically

think about like this a bartebder asjs you what drink your having what is he going to do next ? hes going to make the drink so you can have another right? yha

observe what happens next

the bartender grabs a cup and goes to the ice and scoops it up he takes a bottle of vodka and puts it into the shot cup "a preset cup for alcohol measurement " he takes stawberry puraee (strawberry mix) and pyrs it into the drink and mixes it around by shaking it . puts a extractor on top of it and takes a glass and pours the drink in front of you into the glass. you dont see the vodka in there do you ? no you dont because its mixed in there same thing with lemonade.

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If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma

goldfiish [28.3K]

The balanced chemical reaction:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.

9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3

The limiting reactant is AgNO3.

0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2

0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess

0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess

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Astatine, At, is below iodine in Group 17 of the Periodic Table. Which statement is most likely to be correct?

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The answer is A hope this helps

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A compound consists of atoms of two or more elements combined in a _______.

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Answer: small, whole-number ratio

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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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