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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".

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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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2. HBIO3 name this acid.

aleksandr82 [10.1K]

Answer:

Hydroxy(oxo)bismuthine oxide

3 0

2 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

2 years ago

You need to produce a buffer solution that has a pH of 5.31. You already have a solution that contains 10. mmol (millimoles) of

Karolina [17]

Answer:

37 mmol of acetate need to add to this solution.

Explanation:

Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-

$pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]$

Here pH is 5.31, $pK_{a}$ (acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.

Plug in all the values in the above equation:

$5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]$

or, mmol of $CH_{3}COO^{-}$ = 37

So 37 mmol of acetate need to add to this solution.

3 0

2 years ago

a faucet is leaking water at a speed of 5 drops per minute. If 1 ml is equivalent to 10 drops, how many liters of water will be

nikdorinn [45]

Answer: 0.72 litres of water is wasted in one day.

Explanation:

First you need to find out how many minutes are in a day. Do this by multiplying the number of minutes in an hour (60) by the number of hours in a day (24). 24 x 60 = 1440. If the faucet is dripping at 5 drops per minute, then multiply 5 by the number of minutes in a day (1440) to see how many drops drip in one day. 5 x 1440 = 7200. Now we need to figure out how many mL fo water that is. if 10 drops is 1 mL, then we need to divide the total number of drops (7200) by 10. 7200 divided by 10 is 720. That means 720 mL of water is dripping per day. Finally, we must convert mL to litres. There are 1000 mL in one litre, so divide 720 by 1000. The final answer is 0.72

6 0

2 years ago

What type of physical weathering is caused by the expansion of water?

skelet666 [1.2K]

Answer:

wedging

Explanation:

ice or frost wedging

6 0

2 years ago