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ololo11 [35]
3 years ago
10

Make a generalization about the importance of tools to science.

Chemistry
2 answers:
inessss [21]3 years ago
6 0
Tools are important of science, because of how and when they are used. we use them for many things especially experiments.
nasty-shy [4]3 years ago
4 0
Scientists use special tools to do their work. They gather data, or information, as they seek to learn about the world. In order to solve problems they must record data that is accurate and organized. This is an important part of the scientific method.
You might be interested in
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).
Degger [83]
a) before addition of any KOH : 

when we use the Ka equation & Ka = 4 x 10^-8 : 

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

           = 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

       = 9.2 x 10^-5 M

when PH = -㏒[H+]

   PH = -㏒(9.2 x 10^-5)

        = 4  

b)After addition of 25 mL of KOH: this produces a buffer solution 

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]


first, we have to get moles of HCIO= molarity * volume

                                                           =0.21M * 0.05L

                                                           = 0.0105 moles

then, moles of KOH = molarity * volume 

                                  = 0.21 * 0.025

                                  =0.00525 moles 

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L =  0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

                                        = 0.00525 / 0.075

                                        =0.07 M

and molarity of KCIO = moles KCIO / total volume

                                    = 0.00525 / 0.075

                                    = 0.07 M

and when Ka = 4 x 10^-8 

∴Pka =-㏒Ka

         = -㏒(4 x 10^-8)

         = 7.4 

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4 

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume 

                                                        = 0.21 M * 0.05L

                                                        = 0.0105 moles

then moles KOH = molarity * volume
                            =  0.22 M* 0.035 L 

                            =0.0077 moles 

∴ moles of HCIO remaining = 0.0105 - 0.0077=  8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume 

                                      = 8 x 10^-5 / 0.085

                                      = 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

                                          = 0.0077M / 0.085L

                                          = 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

                                   = 0.0105mol / 0.1 L

                                   = 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

       = 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO] 

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

           = 3.8
∴PH = 14- POH

        =14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

 when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M 

V2 is the excess volume added  of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

           = -㏒0.02 

           = 1.7

∴PH = 14- POH

       = 14- 1.7 

      = 12.3 
8 0
3 years ago
Read 2 more answers
Which radioisotopes have the same decay mode and have half-lives greater than
AveGali [126]

The correct answer is (3)


I-131 and P-32


The explanation:


according to attached table:


- we can see that the half life of p 32 is 14.28d (more than one hour)

- and the half life of I-131 is 8.021 d (more than one hour)


and They both have β- decay mode and with half-lives greater than hour.


4 0
3 years ago
Read 2 more answers
Help, I need to convert 3.16 μm to inches!
irina1246 [14]
0.00011811023622 <span>i</span><span>nches</span>
7 0
3 years ago
Consider the reaction pathway graph below.
Alex17521 [72]
Hello!

The reaction that the graph represents is A. Exothermic because Hrxn=-167 kJ

To calculate Hrxn we apply the following equation:

Hrxn=Hproducts-Hreagents=-625kJ-(-458kJ)=-167kJ

Looking at the graph, and at the result of the calculations, we can see that the enthalpy of the products is lower than the enthalpy of the reagents, because the sign is negative. That means that the reaction releases energy in the form of heat and that the reaction is exothermic.

Have a nice day!
6 0
3 years ago
A weather map indicates that a low-pressure system is approaching a city. Attached to
timofeeve [1]

Answer:

The correct option is;

A warm front is approaching the city and temperatures will increase

Explanation:

A low pressure forms by the movement of air away from a particular spot such that the pressure reduces. The movement of air away from a spot occurs when there is a boundary of warm and cold air such that the air moves to maintain a uniform temperature.

Low pressure centers are associated with warm fronts, resulting in the fall of barometric pressure or observed low pressure and are characterized by a rise in temperature.

3 0
3 years ago
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