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postnew [5]
3 years ago
8

What volume, in milliliters, of 2.0 calcium chloride stock solution would you use to make 500 ml of 0.300 m of calcium chloride

cacl2 solution?
Chemistry
1 answer:
mash [69]3 years ago
8 0
We can use the following formula when making diluted solutions from more concentrated solutions.
c1v1 = c2v2 
where c1 is the concentration and v1 is the volume of the concentrated solution 
and c2 is concentration and v2 is volume of the diluted solution to be prepared 

substituting these values in the equation 
2.0 M x V = 0.300 M x 500 mL 
V = 75 mL

75 mL should be taken from the stock solution and diluted upto 500 mL to make the 0.300 M solution 

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What’s the ionic equation for lithium + water
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Answer:

2Li(s) + 2H₂O(ℓ) ⟶ 2Li⁺(aq) + 2OH⁻(aq) + H₂(g)

Explanation:

An ionic equation uses the symbols (aq) [aqueous] to indicate molecules and ions that are soluble in water, (s) [solid] to indicate insoluble solids, and (ℓ) to indicate substances (usually water) in the liquid state.

In this reaction, solid lithium reacts with liquid water to form soluble lithium hydroxide and gaseous hydrogen .

1. Molecular equation

2Li(s) + 2H₂O(ℓ) ⟶ 2LiOH(aq) + H₂(g)

2. Ionic equation

Lithium hydroxide is a soluble ionic compound, so we write it as hydrated ions.

2Li(s) + 2H₂O(ℓ) ⟶ 2Li⁺(aq) + 2OH⁻(aq) + H₂(g)

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3 years ago
You see the middle layer of the sun’s atmosphere, the _____________, at the start and end of a total eclipse.
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Chromosphere

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When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°
dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.30^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant =  

m= molality = \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579

8.30^0C=1\times K_f\times 0.579

K_f=14.3^0C/m

Let Mass of solute (KBr) = x g

8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}

x=58.2g

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

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2 years ago
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