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3 years ago

What volume is represented by each small tick

Chemistry

1 answer:

sladkih [1.3K]3 years ago

5 0

Answer:

there

Explanation:

Tick Volume Definition Tick volume is measuring every trade whether up or down and the volume that accompanies those trades for a given time period. If you are a day trader or a short term swing trader, tick volume analysis will assist you in sizing up the market on an intraday basis.

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A Solute is the part of the mixture that is<br> A. dissolved<br> B.used to dissolve a substance

8_murik_8 [283]

It is the thing being dissolved!!

5 0

3 years ago

A chemist adds 0.50L of a 0.485 M copper(II) sulfate CuSO4 solution to a reaction flask. Calculate the millimoles of copper(II)

den301095 [7]

Explanation:

It is given that volume is 0.50 L and molarity is 0.485 M. Hence, number of millimoles will be calculated as follows.

Number of millimoles = Molarity × Volume

As there are 1000 mL in 1 L. So, 0.50 L equals 500 mL.

Therefore, putting the given values into the above formula as follows.

Number of millimoles = Molarity × Volume

= 0.485 M × 500 mL

= 242.5

Thus, we can conclude that 242.5 millimoles of copper(II) sulfate has been added by the chemist to the flask.

4 0

3 years ago

In the lab you measure a clean dry crucible and cover to be 24.36 grams. You obtain a 2cm piece of pure magnesium metal. After s

mart [117]

Answer:

1) 0.3g Mg

2)0.5g MgO

3)0.2g O

4)0.01mol Mg & 0.01mol O

5)0.01mol MgO

6) Empirical formula MgO

Explanation:

The mass og Mg is obtained by substracting 24.36g from 24.66g:

24.66 - 24.36 = 0.3g Mg

The ignition of Mg means that it's reacting with oxygen to form an oxide. The increase in the crucible mass after the Mg ignition is due to the addition of oxygen. However, the addition of few drops of water produces a new compound: a hydroxide. According to the oxidation state og Mg (2+), the only magnesium oxide possible is MgO. It happens because the oxidation state of oxygen in oxides is 2-. Which means that just one oxygen atom is required to electrically neutralize one magnesium atom.

We can use a conversion factor to know how much MgO is made from from 0.3 g of Mg:

$0.3g Mg$ * $\frac{16gO}{24.3gMg}$ = 0.2g O

Thereby the mass of the oxide is 0.2g O + 0.3g Mg = 0.5g MgO

We convert the mass of oxygen and magnesium to the respective amounts in moles by using conversion factors:

$0.2g O$ * $\frac{1 mol O}{16g O}$ = 0.01mol O

$0.3g Mg$ * $\frac{1mol Mg}{24.3g Mg}$ = 0.01mol Mg

The moles of MgO can be obtained from:

$0.5g MgO$ * $\frac{1mol MgO}{40.3g MgO}$ = 0.01mol MgO

To obtain the empirical formula, the amount fo moles of each elements must be divided by the smallest one, in this case, 0.01.

The result for both number of Mg atoms and O atoms is 1. This can be interpreted to mean that there is a Mg atom for each O atom forming the formula unit of the compound.

The step when water is added to the compound resulting after heating does not affect the calculations necessary for the magnesium oxide.

4 0

3 years ago

0.22 L of HNO3 is titrated to equivalence using 0.10 L of 0.1 MNaOH. What is the concentration of the HNO3?

White raven [17]

Answer:- Molarity of the acid solution is 0.045M.

Solution:- The balanced equation for the reaction of given acid and base is:

$HNO_3+NaOH\rightarrow NaNO_3+H_2O$

From the balanced equation, they react in 1:1 mol ratio. So, we could easily solve the problem using the equation:

$M_aV_a=M_bV_b$

where, $M_a$ is the molarity of acid, $M_b$ is the molarity of base, $V_a$ is the volume of acid and $V_b$ is the volume of base.

Let's plug in the given values in the equation:

$M_a(0.22L)=0.1M(0.10L)$

on rearranging the above equation:

$M_a=\frac{0.1M(0.10L)}{0.22L}$

$M_a$ = 0.045M

So, the molarity of the acid solution is 0.045M.

3 0

3 years ago

What does NOT affect the characteristic properties of an element?

bazaltina [42]

Answer:

characteristic properties of an element are the defining properties of that element and it does not change with quantity of the element used.

Explanation:

the amount or the quantity of the element used does not affect the characteristic property of the element. it does not matter is the if the amount or the quantity of the element used in the reaction is large or the small the characteristic properties like boiling point, melting point, density, thermal conductivity, etc remain the same or remain constant.

5 0

3 years ago