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Hoochie [10]
3 years ago
9

Environmental studies usually involve an analysis of precipitation and its response to pollution. To quantify the degree of cont

amination in natural rain water or snow, titration is used. The process is quick and results are reliable. Since most titration processes do not require expensive or specialized equipment, the test can be performed often and in different areas with relatively little effort. a. A 1000.0 mL sample of lake water is titrated using 0.100 mL of a 0.100 M base solution. What is the molarity of the acid in the lake water? Show all work. b. Based on the molarity of the acid calculated above, what is the pH of the lake water?
Chemistry
1 answer:
LuckyWell [14K]3 years ago
4 0

The molarity of the lake water is 0.00001 M and the pH of lake water is 5.

The lake water is acidic.

Explanation:

Data given:

molarity of base solution Mbase = 0.1 M

volume of the base solution Vbase = 0.1 ml or 0.0001 litre

volume of lake water Vlake = 1000ml  or 1 litre

molarity of the lake water, Mlake = ?

Using the formula for titration:

Mbase X Vbase = Mlake X

Mlake = \frac{Mbase X Vbase }{Vlake}

Putting the values in the equation:

Mlake = \frac{0.0001 X 0.1}{1}

Mlake = 0.00001 M

The pH of the lake water will be calculated by using the following formula:

pH = - log_{10} [H^{+}]

pH = -log_{10} [ 0.00001]

pH = 5

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<u>Answer:</u> The equilibrium constant for this reaction is 5.85\times 10^{5}

<u>Explanation:</u>

The equation used to calculate standard Gibbs free change is of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]

For the given chemical reaction:

3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

The equation for the standard Gibbs free change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_{(NH_3(g))})]-[(1\times \Delta G^o_{(N_2)})+(3\times \Delta G^o_{(H_2)})]

We are given:

\Delta G^o_{(NH_3(g))}=-16.45kJ/mol\\\Delta G^o_{(N_2)}=0kJ/mol\\\Delta G^o_{(H_2)}=0kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol

To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy = -32.9 kJ/mol = -35900 J/mol  (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 25^oC=[273+25]K=298K

K_{eq} = equilibrium constant at 25°C = ?

Putting values in above equation, we get:

-32900J/mol=-(8.314J/Kmol)\times 298K\times \ln K_{eq}\\\\K_{eq}=e^{13.279}=5.85\times 10^{5}

Hence, the equilibrium constant for this reaction is 5.85\times 10^{5}

5 0
3 years ago
What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen? A. C3H4N B. C2
Ierofanga [76]

Answer:

A. C₃H₄N

Explanation:

  • Firstly, we need to calculate the no. of moles of C, H, and N using the relation:

<em>no. of moles = mass/molar mass.</em>

<em></em>

∴ no. of moles of C = mass/molar mass = (90.0 g)/(12.0 g/mol) = 7.5 mol.

∴ no. of moles of H = mass/molar mass = (11.0 g)/(1.0 g/mol) = 11.0 mol.

∴ no. of moles of N = mass/molar mass = (35.0 g)/(14.0 g/mol) = 2.5 mol.

  • We should get the mole ratio of each atom by dividing by the lowest no. of moles (2.5 mol of N).

∴ the mole ratio of C: H: N = (7.5 mol/2.5 mol): (11.0 mol/2.5 mol): (2.5 mol/2.5 mol) = (3: 4.4: 1) ≅ (3: 4: 1).

  • So, the empirical formula is: A. C₃H₄N.
3 0
4 years ago
Examples of mechanical and physical weathering in our neighborhoods
cestrela7 [59]
When water in a river or stream moves quickly, it can lift up rocks from the bottom of that body of water. When the rocks drop back down they bump into other rocks, and tiny pieces of the rocks break apart
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4 years ago
Is it true dalton inferred that all atoms of an element are identical?
Olegator [25]
Yes and no he said that all atoms of a given element are identical in mass and properties.
7 0
3 years ago
A polymer sample combines five different molecular-weight fractions, each of equal weight. The molecular weights of these fracti
kakasveta [241]

Answer:

Mn = 43,783

Mw = 60,000

Mz = 73,333

narrow distribution = 1.37

Explanation:

The molecular weights of these fractions increase from 20,000 to 100,000 in increments of 20,000. This means their Mi is respectively: (The molar weight (Mi) of the fractions)

Fraction 1 : Mi = 20  *10^3

Fraction 2: Mi = 40 *10^3

Fraction 3 : Mi = 60 *10^3

Fraction 4: Mi = 80 *10^3

Fraction 5 : Mi = 100  *10^3

The ΣMi = 300*10^-3

The Wi (mass of the fractions is for all the fractions the same, let's say 1)

So Wi = 1+1+1+1+1 = 5

Since number of moles = mass / Molar mass

The number of moles is respectively: ni = Wi/Mi (x10^5)

Fraction 1 : ni = Wi/Mi = 1/20000 = 5

Fraction 2: ni = 1/40000 = 2.5

Fraction 3 : ni =1/60000 = 1.67

Fraction 4: ni = 1/80000 = 1.25

Fraction 5 : ni= 1/100000 = 1

The Σni = 11.42

Mn = ΣWi/ni = 5/11.42*10^-5 = 43,783

Mw = (ΣWi * Mi)/ΣWi  = 300,000 /5 = 60,000

Mz = (ΣWi * Mi²)/ΣWi *Mi = (4*10^8 +16*10^8 +36*10^8 +64*10^8 +100*10^8) /300,000  =73,333

Mz/Mn = narrow distribution =60,000/43,783 = 1.37

7 0
3 years ago
Read 2 more answers
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