All categories

3 years ago

Desertification

Chemistry

2 answers:

madam [21]3 years ago

8 0

A.) desertification is when you remove all the trees and most of the plant life in an ares

Reptile [31]3 years ago

6 0

A. will typically increase the amount of erosion in an area
desertification will destroy the land

You might be interested in

During a synthesis reaction, 1.8 grams of magnesium reacted with 6.0 grams of oxygen. What is the maximum amount of magnesium ox

skelet666 [1.2K]

Answer:

2.9 grams.

Explanation:

From the balanced reaction:

Mg + 1/2O₂ → MgO,

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:

no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.

no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.

So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.

Using cross multiplication:

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.074 mol of Mg produce → 0.074 mol of MgO.

∴ The amount of MgO produced = no. of moles x molar mass = (0.074 mol)(40.3 g/mol) = 2.98 g.

5 0

3 years ago

What is meant by Critical temperatue?

yuradex [85]

Answer:

The critical temperature of a substance is the temperature at and above which vapour of the substance cannot be liquefied, no matter how much pressure is applied.

4 0

3 years ago

PLS ANSWER FAST WILL GIVE BRAINLY!!!!

castortr0y [4]

False because the sugar is just changing it’s form

7 0

3 years ago

Read 2 more answers

What is a half-life of an isotope? please help

N76 [4]

Answer:

Explanation:

Hello! Hope this helps!

One important measure of the rate at which a radioactive substance decays is called half-life, or t1/2. Half-life is the amount of time needed for one half of a given quantity of a substance to decay. Half-lives as short as 10–6 second and as long as 109 years are common.

3 0

3 years ago

Please help me like now please

777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

Question 1

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

To balance the equation, count the number of atoms on both sides of the equation

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

Combining the pluses will give you the following

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

Both sides are the same, therefore the chemical equation is balanced (originally).

From the equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.

So that means 0.5 mole of NaHCO3 would produce 0.5 mole of CO2.

Question 2

C4H10 + O2 --> CO2 + H2O

Again, count the number of atoms on both sides of the equation

(4C, 10H, 2O) --> (1C, 2H, 3O) This time left does not equal right side

You now need to find factors that can make both sides equal.

C4H10 + O2 --> 4CO2 + H2O Now the C is balanced, let's recount 

(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H) H&O is still not balanced

C4H10 + O2 --> 4CO2 + 5H2O Now the H is balanced, let's recount

(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H) O is still not balanced

C4H10 + (13/2)O2 --> 4CO2 + 5H2O Now the O is balanced

(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)

But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).

2C4H10 + 13O2 --> 8CO2 + 10H2O Now it's completely balanced!

(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H) Yayy! It's balanced.

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

6 0

2 years ago