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faust18 [17]
3 years ago
13

The maximum allowable concentration of Pb2+ ions in drinking water is 0.05 ppm (i.e., 0.05 g of Pb2+ in 1 million grams of water

). (a) What is the Pb2+ concentration of an underground water supply at equilibrium with the mineral cerussite (PbCO3) (Ksp = 3.3 × 10−14) ? × 10 g / LEnter your answer in scientific notation. (b) Does this concentration exceed the allowable concentration guideline? Yes No
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
6 0

Answer:

a. 0,049g of Pb²⁺ in 1 million grams of water.

b. No

Explanation:

The equilibrium of PbCO₃ in water is:

PbCO₃ ⇄ Pb²⁺ + CO₃²⁻

Where ksp is defined as:

ksp = [Pb²⁺] [CO₃²⁻] <em>(1)</em>

The initial concentration of PbCO₃ in molarity is:

\frac{10g}{1L}×\frac{1mol}{267,21g}= 0,0374M

a. The concentrations in equilibrium are:

[PbCO₃} = 0,0374M-x

[Pb²⁺] = x

[CO₃²⁻] = x

Replacing in (1)

3,3x10⁻¹⁴ = x²

x = 1,82x10⁻⁷M

Thus, [Pb²⁺] = 1,82x10⁻⁷M, this value in g of Pb²⁺ per million grams of water is:

1,82x10⁻⁷M×\frac{267,21g}{1mol}×\frac{1x10^6g}{1000g} = <em>0,049 g of Pb²⁺ in 1 million grams of water</em>

As this value is lower than the maximum allowable concentration of Pb²⁺, <em>the concentration doesn't exceed this maximum value</em>

<em />

I hope it helps!

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With white light, a team measured a 0.7% percent change with 3.5g of plant matter in a one liter container. Convert
Strike441 [17]

moles CO₂ = 5.57.10⁻⁴

<h3>Further explanation   </h3>

A mole is a number of particles(atoms, molecules, ions)  in a substance

Can be formulated :

\tt mol=\dfrac{mass}{MW}

0.7% percent change with 3.5g of plant matter

mass :

\tt 0.7\%\times 3.5~g=0.0245~g

moles :

\tt moles=\dfrac{0.0245}{44}=0.000557=5.57.10^{-4}

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3 years ago
The concentration of CI ion in a sample of H,0 is 15.0 ppm. What mass of CI ion is present in 240.0 mL of H,0, which has a densi
Doss [256]

Answer:

Mass of solute = 0.0036 g

Explanation:

Given data:

Concentration of Cl⁻ = 15.0 ppm

Volume of water = 240 mL

Mass of Cl⁻ present = ?

Solution:

1 mL = 1 g

240 mL = 240 g

Formula:

ppm = mass of solute / mass of sample ×1,000,000

by putting values,

15.0 ppm = (mass of solute / 240 g) ×1,000,000

Mass of solute = 15.0 ppm ×  240 g / 1,000,000

Mass of solute = 0.0036 g

8 0
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An element has an atomic number of 27. How many protons and electrons are in a neutral atom of the element
Sergeu [11.5K]

Answer:

27 and 32

Explanation:

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Which is a component of John Dalton’s atomic theory?
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Atoms of different elements can be identical
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Paul [167]

Answer:

1. Percent composition of  Al = 13.423 %

2.

  • Percent composition of Zn = 28.02 %
  • Percent composition of Cl = 30.6 %
  • Percent composition of O = 41.3 %

3. The empirical formula is C₅O₁₆

4. Molecular Formula= P₄O₆

Explanation:

Part first :

Data Given

Formula of the Molecule = Al₂ (CrO₄)₃

% of Al₂ = ?

> First of all find the atomic masses of each component in a molecule

For Al₂ (CrO₄)₃ atomic masses are given below

Al = 27 g/mol

Cr = 52 g/mol

O = 16 g/mol

> Then find the total masses of each component

2 atoms of Al = 27 g/mol x 2

= 54 g/mol

3 atoms of Cr = 52 g/mol x 3

= 156 g/mol

12 atoms of O = 16 g/mol x 12

= 192 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Al₂ (CrO₄)₃ = [27x2 + 52x3 + 16x12]

Molar Mass of Al₂ (CrO₄)₃ = 402

Now to find the mass percent of Al

Formula used to find the Mass percent of a component

Percent composition of  Al = mass of Al in Molecula / molar mass of Al₂(CrO₄)₃ x 100%

Put the values

Percent composition of  Al =  54 (g/mol) / 402 (g/mol) x 100%

Percent composition of  Al = 13.423 %

_______________________________________

Part 2

Data Given

Formula of the Molecule = Zn(ClO₃)₂

% Zn = ?

% Cl = ?

% O = ?

> First of all find the atomic masses of each component in a molecule

For Zn(ClO₃)₂ atomic masses are given below

Zn = 65 g/mol

Cl = 35.5 g/mol

O = 16 g/mol

> Then find the total masses of each component

1 atoms of Zn= 65 g/mol x 1

= 65 g/mol

2 atoms of Cl = 35.5 g/mol x  

= 71 g/mol

6 atoms of O = 16 g/mol x 6

= 96 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Zn(ClO₃)₂ = [65x1 + 35.5x2 + 16x6]

Molar Mass of Zn(ClO₃)₂ = 232g/mol

Now to find the mass percent of of each component one by one

1.  Formula used to find the mass percent of Zn

Percent composition of  Zn= mass of Zn in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Zn = 65(g/mol) / 232 (g/mol) x 100%

Percent composition of Zn = 28.02 %

-------------------

2.  Formula used to find the mass percent of Cl

Percent composition of  Cl = mass of Cl in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Cl = 71 (g/mol) / 232 (g/mol) x 100%

Percent composition of Cl = 30.6 %

---------------------

3.  Formula used to find the mass percent of O

Percent composition of  O = mass of O in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of O = 96 (g/mol) / 232 (g/mol) x 100%

Percent composition of O = 41.3 %

________________________________________

Part 3:

Data Given

Percentage of C = 27.3 %

Percentage of O = 72.7 %

Emperical Formula of the compound = ?

Solution:

So the compound has 27.3 % C and 72% O

First, find the mass of each of the elements in 100 g of the Compound.

C = 27.3 g

O = 72 g

Now find how many moles are there for each element in 100 g of compound

For this molar mass are required

That is

C = 12 g/mol

O = 16 g/mol

Formula Used

mole of C = mass of C / Molar mass of C

 mole of C = 27.3 / 12 g/mol

  mole of C = 2.275

Formula Used

mole of O = mass of O / Molar mass of O

 mole of O = 72g / 16 g/mol

  mole of O = 7.2

Divide each one by the smallest number of moles

C = 2.275 / 2.275

C = 1

O = 7.2 / 2.275

O = 3.2

Multiply the mole fraction to a number to get the whole number.

C = 1 x 5 = 5

O = 3.2 x 5 =  16

So, the empirical formula is C₅O₁₆

______________________________________

Part 4

Data Given

Percentage of P= 56.38 %

Percentage of O = 43.62%

Molar Mass = 219.9g

Molecular Formula of the compound = ?

Solution:

First, find the mass of each of the elements in 100 g of the Compound.

Mass of P= 56.38g

Mass of O = 43.62g

Now find how many moles are there for each element in 100 g of compound

find the moles in total compounds

Formula Used

mole of P = mass of  / Molar mass of P

 mole of P = 56.38 g / 31 g/mol

  mole of P = 1.818

Formula Used

mole of O = mass of O / Molar mass of O

 mole of O = 43. 62 / 16 g/mol

  mole of O = 2.7262

Now

first find the Emperical formula

Divide each one by the smallest number of moles

P = 1.818 /1.818

P= 1

for oxygen

O = 2.7262 / 1.818

O = 1.5

Multiply the mole fraction to a number to get the whole number.

P = 1 x 2 = 2

O = 1.5 x 2 =  3

So, the empirical formula is P₂O₃

Now  

Find molar mass of the empirical formula P₂O₃

2 (31) + 3 (16) = 62 + 48 = 110

Now find that how many empirical units are in a molecular unit.

(219.9 g/mol) / ( 110 g/mol) =  empirical units per molecular unit

empirical units per molecular unit = 1.999 =2

A here we get two empirical units in a molecular unit,

So the molecular formula is:

2 (P₂O₃) = P₄O₆

7 0
3 years ago
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