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3 years ago

Why calcium oxide and magnesium oxide used as soil treatment

Chemistry

1 answer:

BartSMP [9]3 years ago

4 0

Answer:

Chemically speaking, lime refers only to calcium oxide

(CaO); however, in common usage the term includes the calcination products of calcitic and dolomitic limestones. Calcitic (high-calcium) limes are produced by calcination of

calcareous materials (e.g., calcitic limestone, calcite,

oyster shells, and chalk) containing from 95 to 99 percent

calcium carbonate (CaCO^). Dolomitic limes are produced from

dolomitic limestone or dolomite which contains from 30 to ^+0

percent magnesium carbonate (MgCO^), the rest being calcium

carbonate.

At atmospheric pressure, calcite in limestone decomposes

at approximately 900°C to form CaO and COg. The decomposition of dolomite, CaMg is a two-stage process. At

temperatures between 650°C to 750°C dolomite decomposes to

form MgO, CO^ and CaCO^. It is necessary to raise the temperature to 900°C to decompose the CaCO^ (15, 35)» This

phenomenon is extremely important, as is shown later.

Various investigators have studied the effects of stone

size, temperature, and time of calcination of commercial

Explanation:

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yulyashka [42]

Answer : The image is attached below.

Explanation :

For $O_3$ :

Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n = $\frac{m}{M}=\frac{24g}{48g/mol}=0.5mol$

Number of particles, N = $n\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.0\times 10^{23}$

For $NH_3$ :

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n = $\frac{m}{M}=\frac{170g}{17g/mol}=10mol$

Number of particles, N = $n\times 6.022\times 10^{23}=10\times 6.022\times 10^{23}=6.0\times 10^{24}$

For $F_2$ :

Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n = $\frac{m}{M}=\frac{38g}{38g/mol}=1mol$

Number of particles, N = $n\times 6.022\times 10^{23}=1\times 6.022\times 10^{23}=6.0\times 10^{23}$

For $CO_2$ :

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m = $n\times M=0.10mol\times 44g/mol=4.4g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.10\times 6.022\times 10^{23}=6.0\times 10^{22}$

For $NO_2$ :

Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m = $n\times M=0.20mol\times 46g/mol=9.2g$

Number of particles, N = $n\times 6.022\times 10^{23}=0.20\times 6.022\times 10^{23}=1.2\times 10^{23}$

For $Ne$ :

Molar mass, M = 20 g/mol

Number of particles = $1.5\times 10^{23}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.5\times 10^{23}}{6.022\times 10^{23}}=0.25mol$

Mass, m = $n\times M=0.25mol\times 20g/mol=5g$

For $N_2O$ :

Molar mass, M = 44 g/mol

Number of particles = $1.2\times 10^{24}$

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{1.2\times 10^{24}}{6.022\times 10^{23}}=1.9mol$

Mass, m = $n\times M=1.9mol\times 44g/mol=83.6g$

For unknown substance:

Number of particles = $3.0\times 10^{23}$

Mass, m = 8.5 g

Moles, n = $\frac{N}{6.022\times 10^{23}}=\frac{3.0\times 10^{23}}{6.022\times 10^{23}}=0.50mol$

Molar mass, M = $\frac{m}{n}=\frac{8.5g}{0.50mol}=17g/mol$

The substance is $NH_3$ .

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3 years ago

Hydrochloric acid reacts with sodium hydroxide to produce sodium chloride and water. If 20.6 g of sodium hydroxide reacts with a

astraxan [27]

Answer:

30.1 g NaCl

Explanation:

Your first conversion is converting grams NaOH to moles of NaOH using its molar mass (39.997 g/mol). Then, use the mole ratio of 1 mol NaCl for every 1 mol NaOH to get to moles of NaCl. Then finally multiply by the molar mass of NaCl (58.44 g/mol) to get grams of NaCl.

20.6 g NaOH • (1 mol NaOH / 39.997 g NaOH) • (1 mol NaCl / 1 mol NaOH) • (58.44 g NaCl / 1 mol NaCl) = 30.1 g NaCl

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3 years ago

Why calcium oxide and magnesium oxide used as soil treatment ​

Why calcium oxide and magnesium oxide used as soil treatment