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Archy [21]
2 years ago
15

Describe how to prepare the solution using a 250.0 mL volumetric flask by placing the steps in the correct order. Not all of the

steps will be used.
a. Transfer the measured amount of NaCl to the volumetric flask.
b. Dilute the solution with water to the 250.0 mL mark.
c. Add more water to the flask, but still under a total volume of 250.0 mL, and mix.
d. Hold the cap firmly in place and invert the flask many times.
e. Dissolve the NaCl in less than 250 mL of water and mix well.
f. Dissolve the measured amount of NaCl in 250 mL of water.
Chemistry
1 answer:
padilas [110]2 years ago
6 0

Answer:

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

Explanation:

Preparation of NaCl solution in 250.0 ml volumetric flask:

Add the weighed NaCl directly to volumetric flask and add small amount of water to it and mix it will until all NaCl gets dissolved( if not add small water amount of water more)

After dissolving NaCl add the water upto the mark.

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

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Describe the changes in properties (from metals to nonmetals or from nonmetals to metals) as we move (a) down a periodic group a
defon

Answer:

Explanation:

The changes in properties from metals to non-metals on a periodic table can be measured and determined by the metallicity or electropositivity of elements.

Metallicity is a measure of the tendency of atoms of an element to lose electrons.

a.

Down a periodic group, metallicity increases.

b.

Across a period from left to right electropositivity or metallicity decreases.

Metals are found in the left part of periodic table and the most reactive metal sits in the lower left corner. Non-metals are towards the right side of the table.

7 0
2 years ago
When 200g of AgNO3 solution mixes with 150 g of NaI solution, 2.93 g of AgI precipitates, and the temperature of the solution ri
gizmo_the_mogwai [7]

Answer:

\Delta H=1962.3J

Explanation:

Hello,

In this case, we can compute the change in the solution enthalpy by using the following formula:

\Delta H=mC\Delta T

Whereas the mass of the solution is 350 g, the specific heat capacity is 4.184 J/g °C and the change in the temperature is 1.34 °C, therefore, we obtain:

\Delta H=350g*4.184\frac{J}{g\°C} *1.34\°C\\\\\Delta H=1962.3J

It is important to notice that the mass is just 350 g that is the reacting amount and by means of the law of the conservation of mass, the total mass will remain constant, for that reason we compute the change in the enthalpy as shown above, which is positive due to the temperature raise.

Best regards.

8 0
2 years ago
Balance the equation xcl2(aq)+agno3(aq)=x(no3)2(aq)+agcl(s)
lisabon 2012 [21]

Explanation:

XCl _{2(aq)} + 2AgNO _{3(aq)}→X(NO _{3}) _{2(aq)}   +2 AgCl _{(s)}

6 0
2 years ago
Read 2 more answers
Magnesium and nitrogen react in a combination reaction to produce magnesium nitride: 3 Mg N2 → Mg3N2 In a particular experiment,
vlada-n [284]

Answer:

The mass of Mg consumed is 21.42g

Explanation:

The reaction is

3Mg+N_{2}-->Mg_{3}N_{2}

As per balanced equation, three moles of Mg will react with one mole of nitrogen to give one mole of magnesium nitride.

as given that mass of nitrogen reacted = 8.33g

So moles of nitrogen reacted = \frac{mass}{molarmass}=\frac{8.33}{28}=0.2975mol

moles of Mg required = 3 X moles of nitrogen taken = 3X0.2975 = 0.8925mol

Mass of Mg required = moles X molar mass = 0.8925 X 24 = 21.42 g

5 0
2 years ago
7.744 Liters of nitrogen are contained in a container. Convert this amount to grams.
Xelga [282]

Answer:

9.69g

Explanation:

To obtain the desired result, first let us calculate the number of mole of N2 in 7.744L of the gas.

1mole of a gas occupies 22.4L at stp.

Therefore, Xmol of nitrogen gas(N2) will occupy 7.744L i.e

Xmol of N2 = 7.744/22.4 = 0.346 mole

Now let us convert 0.346 mole of N2 to gram in order to obtain the desired result. This is illustrated below:

Molar Mass of N2 = 2x14 = 28g/mol

Number of mole N2 = 0.346 mole

Mass of N2 =?

Mass = number of mole x molar Mass

Mass of N2 = 0.346 x 28

Mass of N2 = 9.69g

Therefore, 7.744L of N2 contains 9.69g of N2

7 0
2 years ago
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