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2 years ago

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Chemistry

2 answers:

DIA [1.3K]2 years ago

7 0

Explanation:

Ground water is defined as the water present below the ground. When chemicals spill from factories, industries etc then they are absorbed by soil from where it flows through the pores present in it.

Pesticides and fertilizers also contain toxic substances which flow and absorbed by the ground when it rains.

Hence, they also contribute to water pollution.

Thus, we can conclude that ground water pollution can be caused by chemical spills, pesticides and fertilizers.

Vesna [10]2 years ago

3 0

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It's All of the above.

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What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0

expeople1 [14]

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = 45.1mL

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = $\frac{[OH^-][HNO_2]}{[NO_2]}$

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

Where X is reaction coordinate

Replacing in Kb:

1.41x10⁻¹¹ = $\frac{[X][X]}{[0.211 -X]}$

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ FALSE ANSWER. There is no negative concentrations.

X = 1.72x10⁻⁶. Right answer.

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; pH = 8.24

3 0

3 years ago

Aluminum foil is often incorrectly termed tin foil. If the density of tin is 7.28g/cm³,what is the thickness of a piece of tin f

leonid [27]

Answer:

15.0 µm

Step-by-step explanation:

Density = mass/volume

D = m/V Multiply each side by V

DV = m Divide each side by D

V = m/D

Data:

m = 1.091 g

D = 7.28 g/cm³

l = 10.0 cm

w = 10.0 cm

Calculation:

(a) Volume of foil

V = 1.091 g × (1 cm³/7.28 g)

= 0.1499 cm³

(b) Thickness of foil

The foil is a rectangular solid.

V = lwh Divide each side by lw

h = V/(lw)

= 0.1499/(10 × 10)

= 1.50 × 10⁻³ cm Convert to millimetres

= 0.015 mm Convert to micrometres

= 15.0 µm

The foil is 15.0 µm thick.

4 0

2 years ago

Calculate the [H3O+] of solutions a and b; calculate the [OH-] solutions c and d.

patriot [66]

Answer:

I got the answers but it won't let me post it correctly on here....

Explanation:

9.) 10-2.76 =0.0174 [H30+]= 1.74*10-3 M

10.)10-3.65=0.00224  [H3O+] =2.24*10-2 M

11.)10-3.65=0.00224 [OH-]= 2.224*10-4M

12.)10-6.87=0.00000135  [OH-]= 1.35*10-7M

3 0

2 years ago

please hurry! If 3.87g of powdered aluminum oxide is placed in a container containing 5.67g of water, what is the limiting react

VikaD [51]

Answer:

Explanation:

Given parameters:

Mass of aluminium oxide = 3.87g

Mass of water = 5.67g

Unknown:

Limiting reactant = ?

Solution:

The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.

Using the number of moles, we can ascertain the limiting reactants;

Al₂O₃ + 3H₂O → 2Al(OH)₃

Number of moles;

Number of moles = $\frac{mass}{molar mass}$

molar mass of Al₂O₃ = (2x27) + 3(16) = 102g/mole

number of moles = $\frac{3.87}{102}$ = 0.04mole

molar mass of H₂O = 2(1) + 16 = 18g/mole

number of moles = $\frac{5.67}{18}$ = 0.32mole

From the reaction equation;

1 mole of Al₂O₃ reacted with 3 moles of H₂O

0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O

But we were given 0.32 mole of H₂O and this is in excess of amount required.

This shows that Al₂O₃ is the limiting reactant

6 0

2 years ago

Which of the following is a property of matter?

lana66690 [7]

Explanation:

Matter also exhibits physical properties. Physical properties are used to observe and describe matter. Physical properties can be observed or measured without changing the composition of matter. These are properties such as mass, weight, volume, and density.

5 0

2 years ago