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3 years ago

Whether an element is classified as a major or as a trace mineral is determined by

2 answers:

5 0

Whether an element is classified as a major or as a trace mineral is determined by "<span>the amounts daily required by the body".

</span>

Calcium, magnesium, phosphorus, potassium, chloride and sulfur are all examples of the major minerals and these are the minerals a body needs in large quantity as opposed to include iron, copper, zinc, manganese, fluoride, selenium and cobalt, which are all trace minerals and are needed in very little quantity.

Svetllana [295]3 years ago

4 0

<h2>Element Trace Mineral is Determination</h2>

Whether an element is classified as a major or as a trace mineral is determined by the amounts needed daily by the body. Species need minerals in varying amounts, but commonly required plant trace elements such as copper (Cu), Boron (B), Zinc (Zn), Manganese (Mn) and Molybdenum (Mo).

Animals require Manganese (Mn), iodine (I) and cobalt (Co). Human requires trace minerals in small amounts such as iron (Fe), iodine (I), fluoride (F), copper (Cu), Zinc (Zn), Chromium (Cr), Selenium (Se), Manganese (Mn) and Molybdenum (Mo).

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In the following reaction, how many liters of oxygen will react with 270 liters of ethene (C2H4) at STP?

lawyer [7]

According to the equation you have 1 mole of C2H4 and 3 moles of O2.

1 • (22.4L / 270L) = 3 • (22.4L / x)

1/270L = 3/x

x = 3(270) / 1

x = 810 L

810 Liters of oxygen will react with 270 liters of ethene (C2H4) at STP

4 0

2 years ago

What is the Ka value if pka=10.36

siniylev [52]

Pka = - log ka

10.36 = - log ka

Ka = .......

5 0

3 years ago

Read 2 more answers

What is the percent yield of a reaction in which 51.5 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce m

Rus_ich [418]

Answer:

The percent yield of a reaction is 48.05%.

Explanation:

$WO_3+3H_2\rightarrow W+3H_2O$

Volume of water obtained from the reaction , V= 5.76 mL

Mass of water = m = Experimental yield of water

Density of water = d = 1.00 g/mL

$M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g$

Theoretical yield of water : T

Moles of tungsten(VI) oxide = $\frac{51.5 g}{232 g/mol}=0.2220 mol$

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:

$\frac{3}{1}\times 0.2220 mol=0.6660 mol$

Mass of 0.6660 moles of water:

0.666 mol × 18 g/mol = 11.988 g

Theoretical yield of water : T = 11.988 g

To calculate the percentage yield of reaction , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%$

The percent yield of a reaction is 48.05%.

7 0

3 years ago

Read 2 more answers

there is a total of seventeen constitutional isomers for the molecular formula c5h13n. draw the skeletal formula of all three co

MAXImum [283]

Amines are derivatives of Ammonia (NH₃) in which atleast one hydrogen atom is replaced by an alkyl group. Amines are further classifies as;

Primary Amines:
In primary amines the nitrogen atom is attached to two hydrogen atoms and one alkyl group.

Secondary Amines:
In secondary amines the nitrogen atom is attached to two alkyl groups and one hydrogen atom.

Tertiary Amines:
In tertiary amines the nitrogen atom is attached to three alkyl groups, hence it has no hydrogen atom.

Below are three isomers of tertiary amines with molecular formula C₅H₁₃N.

5 0

3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$