NaOH + HCl → NaCl + H2O An instructor is planning to demonstrate this reaction to a class as an example of a neutralization reaction. She does a test run, using NaOH pellets and 1.0 M HCl. The reaction mixture spatters. She concludes that this is unsafe; the reaction rate is too fast. How can she slow it down?

heat the HCl

use 0.2 M HCl

add a catalyst

grind the NaOH pellets to a powder


Answer 1

The correct answer is B).

Answer 2


B) use 0.2 M HCl


Lowering the concentration of a reactant will slow down the rate of reaction. If she would use 0.2 M HCl, the reaction rate will be slower than with 1.0 M HCl. The other choices would increase the rate of reaction, making it even more dangerous.

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Please Help! A drug company is testing the effectiveness of a new blood pressure medicine using ratstas the test subjects.

a.Describe the experimental group:

b.Describe the control group:

c.what is the independent variable?

d.what is the dependent variable?



Answer: a. Group of rats which was tested.

b. Group of rats which was left untreated.

c. The new blood pressure medicine.

d. Effectiveness of the blood pressure medicine on the rats.


In an experimental setup certain variables are taken which are examined to determine the result of the experiment. The independent variable can be manipulated in an experiment so as to determine it's influence over the experimental group. The experimental group is the one which is being treated in an experiment. The control group is the one which is not treated in an experiment. It is kept as a standard to observe and compare the changes occurring in the experimental variable. The outcome of the changes occurring in the experimental variable can be used to determine the result of the experiment, in form of dependent variable.

a. Group of rats which was tested.: One group of rats can be used to test the effectiveness of the medicine will be included as experimental group.

b. Group of rats which was left untreated.: One group of rats can be used as a control group which is left undisturbed without treatment.

c. The new blood pressure medicine.: The new blood pressure medicine is the independent variable as it's affect and efficiency is undetermined for blood pressure.

d. Effectiveness of the blood pressure medicine on the rats.: The outcome of drug testing should be the determination of the effectiveness of the medicine on the blood pressure of rats. Hence, this is the dependent variable.

A. group of rats being tested
b. group of rats not being tested
c. independent variable is what you change (blood pressure medicine)
d. dependent variable depends on what the independent variable does (outcome of testing the rats)

If an atom has 1 to 3 valence electrons, what will it do? -move electrons to a lower shell

-accept electrons

-move electrons to a higher shell

-donate electrons




An atom with one to three valence electrons would have to gain seven to five valence electrons, respectively to get a stable octet.

It is easier for it to donate electrons and expose the stable octet of electrons in the shell below..

(A) is wrong. You can't move electrons to a lower shell, because it's already full.

(B) is wrong. The atom will donate electrons.

(C) is wrong. Moving electrons to a higher shell will make the atom more unstable.


48.0 mL of 1.70 M CuCl2(aq) and 57.0 mL of 0.800 M (NH4)2S(aq) are mixed together to give CuS(s) as a precipitate. The other product of the reaction is aqueous ammonium chloride. What is the concentration of the Cu(II) ion after the complete reaction?



The concentration of the Cu(II) ion is 0.777M


Step 1: Data given

Volume of  1.70 M CuCl2 = 48.0 mL = 0.0480 L

Volume of 0.800 M (NH4)2S = 57.0 mL = 0.0570 L

Step 2: The balanced equation

CuCl2 (aq) + (NH4)2S (aq) → 2 NH4Cl (aq) + CuS (s)

Step 3: Calculate moles CuCl2

moles CuCl2 = 0.0480 L * 1.70 M=0.0816 moles

Step 4: Calculate moles (NH4)2S

moles (NH4)2S = 0.0570 L * 0.800 M = 0.0456 moles

Step 5: Calculate the limiting reactant

The ratio between CuCl2 and (NH4)2S is 1 : 1 so (NH4)2S is the limiting reactant . IT will completely be consumed (0.0456 moles).

CuCl2 is in excess. There will remain 0.0816 - 0.0456 = 0.0360 moles

Step 6: Calculate moles of CuS

For 1 mol CuCl2 we need 1 mol (NH4)2S to produce 2 moles of NH4Cl and 1 mol CuS

For 0.0456 moles we'll produce 0.0456 moles CuS

Step 7: Calculate moles of Cu(II)ion

There remains 0.0360 moles CuCl2.

There will be 0.0456 moles CuS produced

Total moles Cu^2+ = 0.0816 moles

Step 8: Calculate concentration of Cu(II) ion

Concentration = moles / volume

Concentration = 0.0816 moles / (0.048+0.057)

Concentration = 0.777 M

The concentration of the Cu(II) ion is 0.777M


‘A’ is an element which belongs to period 3, having 6 electrons in its valence shell. Below is a list of successive ionization energies (in kJ/mol) for period 3. IE2 = 2250 IE3 = 3360 IE4 = 4560 IE5 = 7010 IE6 = 8500 IE7 = 27,100 Identify the element ‘A’



The element A is S (sulfur)


The elements for the 3erd period in the periodic table are Na, Mg, Al, Si, P, S, Cl and Ar.

The one that has 6 e⁻ in its valence shell is the S, because it is missing 2 e⁻ to reach the octet rule. 2 e⁻ to has the most stable noble gas conformation.

The IE of S = 3360 kJ/mol

It is a little lower than Cl because the electron is so far from the nucleus, that's why we have to apply a very low ionization energy to rip the electron off.

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