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Shalnov [3]
3 years ago
9

A solution of salt and water is 33.0% salt by mass and has a density of 1.50 g/ml. what mass of the salt in grams is in 5.00l of

this solution?
Chemistry
1 answer:
meriva3 years ago
6 0
To answer this item, we solve first for the mass of the solution by multiplying the density by the volume. That is,

      m = (density)(volume)

Substituting the known values,
    m = (1.50 g/mL)(5L)(1000 mL/1L)
      m = 7500 grams

To determine the mass of the salt in the solution, multiply the calculated mass of the solution by the decimal equivalent of the percent salt in the solution.

      m of salt = (7500 g)(0.33)
      m of salt = 2475 grams

<em>Answer: 2475 grams</em>
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Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
barxatty [35]

Answer:

[Pb^{2+}]=3.9 \times 10^{-2}M

this is the concentration required to initiate precipitation

Explanation:

PbCl_2  ⇄ Pb^{2+}+2Cl^-

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

Ksp=2.4\times 10^{-4}

lets solubility=S

[Pb^{2+}] = S

[Cl^-]=2S

Ksp=[Pb^{2+}]\times [Cl^-]^2

Ksp=S \times (2S)^2

Ksp=4S^3

S=\sqrt[3]{\frac{Ksp}{4} }

S=3.9\times 10^{-2}

[Pb^{2+}]=3.9 \times 10^{-2}M this is the concentration required to initiate precipitation

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What type of air masses might make up the cold air mass in a cold front
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Calculate the number of O atoms in 0.364 g of CaSO4 · 2H2O
Nikolay [14]

Answer:

<em>= 7.66 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

Explanation:

For problems like this posting, one needs an understanding of the following topics:

The definition of the mole

<u>1 mole of substance</u> = mass in grams of substance containing 1 Avogadro's Number ( = 6.023 x 10²³ ) of particles of the specified substance. This is generally one formula weight of the substance of interest. From this, the following equivalent relationships should be memorized:

<em>   1 mole = 1 formula weight = 1 mole weight (g)= 6.023 x 10²³ particles</em>

Converting grams to moles:

<em>Given grams => moles = grams/gram formula wt </em>

Converting moles to grams:

<em>Given moles => grams = moles x gram formula wt</em>

_________________________________________________________

<em>Calculate the number of O atoms in 0.364 g of CaSO₄ · 2H₂O.</em>

Given mass CaSO₄ · 2H₂O = 0.364 grams

Formula Wt CaSO₄ · 2H₂O = 172 g/mole

moles CaSO₄ · 2H₂O = mass <em>CaSO4 · 2H2O / formula Wt. CaSO₄ · 2H₂O</em>

<em>= 0.364 g CaSO₄·2H₂O </em><em>/ </em><em>172 g CaSO4·2H2O </em>

<em>= (0.364/172) mole CaSO₄·2H₂O </em>

<em>= 2.12 x 10⁻³ mole CaSO₄·2H₂O    </em>

<em>∴ number of Oxy (O) atoms in 0.364 grams CaSO₄·2H₂O </em>

<em>=  (2.12 x 10⁻³ mole CaSO₄ · 2H₂O)(6.023 x 10²³ molecules CaSO₄· 2H₂O/ mole)</em>

<em>= 1.276876 x 10²¹molecules CaSO₄· 2H₂O  CaSO₄2H₂O </em>

<em>= 1.276876 x 10²¹ molecules CaSO₄· 2H₂O   x   6 oxygen atoms / molecule</em>

<em>= 7.661256 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

<em>= 7.66 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

<em />

8 0
2 years ago
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