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3 years ago

What % of the ideal soil is air?

Chemistry

1 answer:

alina1380 [7]3 years ago

4 0

Answer:

In soil air as in the atmosphere, nitrogen gas (dinitrogen) comprises about 78%. In the atmosphere, oxygen comprises about 21% and carbon dioxide comprises about 0.36%. However, in the soil air, oxygen usually is replaced by carbon dioxide, so both range from about 0.4% to 21%.

Explanation:

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Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro

melamori03 [73]

Answer: The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For nickel:

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol$

For the given chemical reaction:

$3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)$

For nickel (II) oxide:

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from $\frac{3}{3}\times 0.252=0.252mol$ of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

$0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g$

For aluminium:

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from $\frac{2}{3}\times 0.252=0.168mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

$0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g$

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

4 0

3 years ago

What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, c2h6? in this reaction,

Vitek1552 [10]

Balance equation for combustion of ethane will be:

2C₂H₆(g) + 7O₂(g)--------> 4CO₂(g) + 6H₂O(g)

To balance the equation:

1. Balance the number of carbon atom on both side:

C₂H₆(g) + O₂(g)--------> CO₂(g) + H₂O(g)

1. balance the number of carbon on both side, as in reactant there are 2 but in product one,

so , multiply the CO₂, by 2 in the product.

2. Balance the number of hydrogen on both side as in reactant the number of hydrogen is 3 but in product it is 6 so, multiplythe number of H₂O by 3,

so multiply the number of H₂O by 3 in product.

3. Balance the number of oxygen on both side , as 1 and 2 step increases the number of oxygen and it becomes 7 , so to balance the number of oxygen on both side by mutiplying the number of O₂ by 7/2 in reactant .

4. Now, doubling the equation will give balance equation that is:

2C₂H₆(g) + 7O₂(g)--------> 4CO₂(g) + 6H₂O(g)

8 0

3 years ago

19 The coefficients in a balanced chemical equation represent(l ) the mass ratios of the substances in the reaction(2) the mole

Andrews [41]

2. The coefficients represent to molar ratios in a balanced equation.

8 0

3 years ago

True or False? Nuclear fusion releases less energy than nuclear fission does.

elixir [45]

False

nuclear fusion produces more energy than a nuclear fission reaction.

8 0

2 years ago

Read 2 more answers

Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$