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3 years ago

7

Farmer Jones has a problem with his field. Acid rain has made it too acidic. What type of substance should he add to neutralise

the acid?

1 answer:

Crank3 years ago

7 0

Answer:

calcium oxide to nutriolise the acidity of the soil

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State three reasons why an airborne pollutant might be deposited close to its source

den301095 [7]

Explanation:

Air borne pollutants are usually very light particles that are usually carried in the air in the atmosphere. Some of these pollutants can be carried far away and some are found close to their sources.

Some of the factors are:

Wind pressure: if wind pressure is low, pollutants will not be carried far away. In this situation, the particles will be blown and carried over a short distance from their sources. When wind pressure is very high, pollutants will be carried extensively into farther areas.
Pollutants can be trapped in snow. In a snowy area, when pollutants reach sky high and enters condensed water in the atmosphere, they can become super cooled into ice and fall back within the same area where they are formed.
Rain water can also cause pollutants to be dispersed within a short area. Pollutants can mix with rain water and carrier back to the area where they were originally formed.

Learn more:

Pollution brainly.com/question/10743354

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3 0

3 years ago

The ksp of cadmium hydroxide, cd(oh)2, is 7.20 × 10-15. calculate the solubility of this compound in g/l.

olga nikolaevna [1]

When the solubility is the amount of solute that dissolved in a unit volume of the solvent.

and when the balanced reaction equation is:

so, by using the ICE table:

Cd(OH)2(s) ↔ Cd2+ + 2OH-

initial 0 0

change +X +2X

Equ X 2X

so the Ksp expression = [Cd2+] [OH-]^2

and we have Ksp = 7.2 x 10^-15

and when we assume the solubility = X

by substitution:

7.2 x 10^-15 = (X) (2X)^2

7.2 x 10^-15 = X*4X^2

7.2 x 10^-15 = 4X^3

∴X = ∛(7.2x10^-15)/4

= 1.22 x 10^-5 M

∴ the solubility = X = 1.22 x 10^-5 M

6 0

4 years ago

If the relative errors of three quantities to be multiplied together are 0.009, 0.006 and 0.003, what is the relative error of t

Dima020 [189]

Answer: The relative error of the resulting quantity is 0.018.

Explanation: Relative error of the quantities when are multiplied together is usually less than or equal to the sum of each relative error. Mathematically, it is represented as

$\delta x=\frac{\Delta x}{x}$

According to the question,

Let us assume that the three quantities are $r_1,r_2\text{ and }r_3$

$r=r_1\times r_2\times r_3$

Taking log on both the sides, we get

$log(r)=log(r_1\times r_2\times r_3)$

$log(r)=log(r_1)+log(r_2)+log(r_3)$

Relative error is calculated by:

$\frac{\delta r}{r}=\frac{\delta r_1}{r_1}+\frac{\delta r_2}{r_2}+\frac{\delta r_3}{r_3}$

$\frac{\delta r}{r}=0.009+0.006+0.003=0.018$

This value has 3 significant figures only, so the resulting relative error is 0.018.

4 0

4 years ago

Dacă intr-o scoica se introduc câteva picături de acid se observa apariția efervescenței. Scoica conține carbonat de calciu.Daca

Step2247 [10]

Answer:

Explanation:

In English;

If a few drops of acid are introduced into a shell, the effervescence may be observed. The shell contains calcium carbonate.If 1.46g of hydrochloric acid have been introduced, what quantities of reaction products are obtained?

Given parameters:

Mass of HCl = 1.46g

Unknown:

Mass of CO₂ produced = ?

Mass of H₂O produced =?

Mass of CaCl₂ produced = ?

Solution

The balanced reaction equation is given below:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

The effervescence observed is due to the production of carbon dioxide gas.

In order to find the mass of each of the products obtain, we must first express the known mass of the acid in terms of moles:

number of moles of HCl = $\frac{mass of HCl}{molar mass of HCl}$

Molar mass of HCl = 1 + 35.5 = 36.5g/mol

number of moles of HCl = $\frac{1.46}{36.5}$ = 0.04mole

Now, we must find the number of moles of each of the products obtained

number of moles of CaCl₂:

2 moles of HCl produced 1 mole of CaCl₂

0.04 moles of HCl will produce $\frac{0.04}{2}$ moles = 0.02moles of CaCl₂

Also, number of mole of H₂O is 0.02moles

number of mole of CO₂ is 0.02moles

Mass of the products:

Mass of CaCl₂ = number of moles of CaCl₂ x molar mass of CaCl₂

molar mass of CaCl₂ = 40 + 2(35.5) = 111g/mol

mass of CaCl₂ = 0.02 x 111 = 2.22g

Mass of H₂O = number of moles of H₂O x molar mass of H₂O

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

mass of H₂O = 0.02 x 18 = 0.36g

Mass of CO₂ = number of moles of CO₂ x molar mass of CO₂

molar mass of CO₂ = 12 + 2(16) = 44g/mol

mass of CO₂ = 0.02 x 44 = 0.88g

5 0

4 years ago

How does the structure of the cell membrane help the cell regulate water?

Maslowich

By a process called osmosis. osmosis depends on the concentration of the water inside and outside the cell

6 0

3 years ago