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Featured 3 months ago

There isn't a single tool. We have many different glassware that we can use, and some have greater precision (number of known digits, plus one guessed decimal place) than others.

Typical glassware that can measure liquid volume are:

**Graduated Cylinders**(generally uncertainties are in the tenths place, since the index markings are at each integer)

**Erlenmeyer Flasks**(generally uncertainties are fairly high depending on the flask volume, so this is mainly a reaction vessel but can be used to approximate volume)

**Volumetric Flasks**(often calibrated for containing or delivering a*specific*volume, great for measuring exact volumes. These often have uncertainties in the*hundredths*place, such as#100.00 pm "0.02 mL"# , when you properly measure to the index mark.)

**Burettes**(fairly common; used in titrations fairly often, with an uncertainty usually in the hundredths place. For example,#100.00 pm "0.10 mL"# for a to-deliver burette.)

**Graduated Pipettes**(not too common, but you might see these. I used these several times, and kept a collection of them throughout my quantitative classes. Requires a bulb.)

**Volumetric Pipettes**(hard to get right for undergrad students, but they are often made to deliver exactly to the hundredths place or so, when properly measured to the index marking. For example, you can get#10.00 pm "0.01 mL"# on a Class A [to deliver] volumetric pipette.)

As chemists, we have all these tools (and more) to measure liquid volume, depending on the situation.

- If we only need to
**estimate**or get**rough**quantities,*graduated cylinders*and*Erlenmeyer flasks*are OK for the job. - Often in
**quantitative**situations,*volumetric pipettes*and*volumetric flasks*are preferred, particularly when you are trying to measure a very specific volume. *Graduated pipettes*are used as well in**quantitative**situations sometimes. At least, I used them quite a bit.

Featured 5 months ago

The basic formula is

To answer this, we can take a look at the **ionic charges** of the elements involved.

Since oxygen is in group 16, it will have

Recall that an atom will want to gain or lose electrons to fulfill the *octet rule*; it wants to do whatever is easiest to obtain

For an oxygen atom, the easiest way to get

For the alkali metals, which are located in group 1 of the periodic table, they each have

In order to obtain an octet for each of the alkali metals, the easiest way to do it is by removing its one valence electron. This will cause each of the atoms (symbol

**The formula**:

The compound the alkali metal *electrically neutral*. That is to say, the charges must balance out to

The easiest way to write a formula of any ionic compound if you know the ionic charges of each species is to **put the charge of the anion (the negative ion, in this case is #"O"^(2-)#), #2-# as the ***. Here's what I mean, using sodium (

Featured 4 months ago

0.980 g

The ratio of the amounts of U 238 and Pb 206 in a rock sample enables the age of the rock to be estimated using the technique of radiometric dating.

U 238 forms a decay chain in which it undergoes a sequence of 8

It moves back in the periodic table until the isotope falls in the band of stability at Pb 206.

Each step has its own individual half - life but the first decay to Th 234 is about 20,000 times slower than the other decay steps.

Those of you who are familiar with chemical kinetics will know that it is the slowest step in a mechanism which determines the overall rate of reaction, the so - called "rate determining step".

This is the case here in the conversion of U 238 to Pb 206.

As the U 238 decays exponentially, the amount of Pb 206 grows correspondingly:

The half - life of U 238 is about 4.5 billion years. As time passes, the ratio of Pb 206 to U 238 will increase and it is this which enables the age of the rock to be estimated.

In this question, we are given the age of the sample which we can use to estimate the amount of Pb 206.

We need to do some "math" to show this:

Radioactive decay is a first order process:

Since the decay of 1 U 238 atom will result in the formation of 1 atom of Pb 206 we can say that:

Where

The decay equation can therefore be written:

The half - life,

We can get the value of the decay constant from the expression:

We can get the number of moles of

We don't know the mass of Pb 206 so we will call this **"m"** :

There is no need to convert these into a number of atoms by multiplying by the Avogadro Constant, since we are interested in the Pb 206 / U 238 ratio so this would cancel anyway.

Taking natural logs of both sides

Putting in the values for

Featured 3 months ago

see explanation => ...

The following is for posted equation ...

Featured 1 month ago

#t = "145.5 s"#

(For the record, I did this without looking at the answer. It's much more important you know HOW to do it.)

The first method is a very general approach as in radioactive decay problems (which are first-order processes!). The second method is from the kinetics unit in general chemistry.

**METHOD 1**

This can be done simply by knowing the definition of a half-life:

The time it takes for the concentration of a sample to halve.

Since we know that

#t_"1/2" = "43.80 s"# ,

consider the following recursive train of thought **for first-order half-lives only**.

- After one half-lives,
#[A] = 1/2[A]_0# . - After two half-lives,
#[A] = 1/4[A]_0# . - After three half-lives,
#[A] = 1/8[A]_0# . - After four half-lives,
#[A] = 1/16[A]_0# .

and so on. Thus, we can write the **current concentration** as a function of the starting concentration:

#[A] = (1/2)^n [A]_0# where

#n# is the number of half-lives that passed.

The **number of half-lives passed**,

Thus,

#barul|stackrel(" ")(" "[A] = (1/2)^(t//t_"1/2") [A]_0" ")|#

Given that

#[A] = 0.1[A]_0# ,

and thus,

#(1/2)^(t//t_"1/2") = 0.1# .

To solve for

#ln(1/2)^(t//t_"1//2") = ln0.1#

#t/(t_"1/2")ln(1/2) = ln0.1#

Therefore, it will take this long for

#color(blue)(t) = t_"1/2"cdotln0.1/ln(1/2)#

#= "43.80 s" cdot (ln 0.1)/(ln (1/2))#

#=# #color(blue)("145.5 s")#

**METHOD 2**

The more usual method uses the **half-life of a first-order reaction** (refer to your textbook):

#t_"1/2" = (ln2)/k#

And thus, the rate constant is given by:

#k = (ln2)/t_"1/2"#

Via the **first-order integrated rate law** (refer to your textbook),

#ln[A] = -kt + ln[A]_0# ,

one can then solve for the time passed simply by knowing that

#ln(0.1[A]_0) - ln[A]_0 = -kt#

#ln((0.1cancel([A]_0))/(cancel([A]_0))) = -kt#

#t = ln0.1/(-k)#

#= (ln0.1)/(- (ln2)/(t_"1/2"))#

#= t_"1/2" cdot (ln0.1)/(ln (1/2))#

which is of the same form as in method 1. We would also get

Featured 1 month ago

#DeltaT_f ~~ -0.015^@ "C"#

This change was so small since the solid-liquid coexistence curve is often very steep. So the small pressure change typically doesn't alter the melting point that much.

A few things to keep in order:

- Seems like there's a typo. There are
#10^10 "ergs"# in#"1 kJ"# , so you should have had#DeltaH_"fus" = 796 * 4.186 * 10^6 "erg/g"# , or#"6.02 kJ/mol"# . - The specific volume of a substance just an old way to specify a reciprocal density. Not that it matters, because we know that the densities should be close to
#"1 g/cm"^3# ... and they are:

#rho_w = 1/1.0001 "g/cm"^3 = "0.9999 g/cm"^3#

#rho_"ice" = 1/1.0908 "g/cm"^3 = "0.9168 g/cm"^3#

Now, since you have to examine the change in melting point with pressure variance, consider the **Clapeyron equation**:

#(dP)/(dT) = (DeltabarH_"fus")/(T_fDeltabarV_"fus")# where:

#(dP)/(dT)# is theslope of the two-phase coexistence curveon a phase diagram.#DeltabarH_"fus"# is thechange in molar enthalpy of fusionin#"kJ/mol"# .#T_f# is thefreezing pointof the substance in#"K"# .#DeltabarV_"fus" = barV_w - barV_"ice"# is thechange in molar volumein#"L/mol"# due to melting ice.

We already converted

#barV_w = [(0.9999 cancel("g H"_2"O"))/cancel("cm"^3) xx cancel("1 cm"^3)/cancel"mL" xx (1000 cancel"mL")/"L" xx ("1 mol")/(18.015 cancel("g H"_2"O"))]^(-1)#

#= 1/("55.504 mol/L") = "0.01802 L/mol"#

#barV_"ice" = [(0.9168 cancel("g H"_2"O"))/cancel("cm"^3) xx cancel("1 cm"^3)/cancel"mL" xx (1000 cancel"mL")/"L" xx ("1 mol")/(18.015 cancel("g H"_2"O"))]^(-1)#

#= 1/("50.891 mol/L") = "0.01965 L/mol"#

Now, to calculate the change in melting point, we consider the slope given by

In some small interval where the temperature and pressure change, we consider the reference temperature and pressure to be **normal melting point**:

#(dP)/(dT_f) ~~ (DeltaP)/(DeltaT_f) = ("3 atm" - "1 atm")/(T_f' - "273.15 K")#

The right-hand side of the equation becomes:

#(DeltaP)/(DeltaT_f) ~~ (6.02 cancel"kJ""/"cancel"mol")/("273.15 K" cdot (0.01802 cancel"L""/"cancel"mol" - 0.01965 cancel"L""/"cancel"mol")) xx (0.082057 cancel"L"cdot"atm")/(8.314472 xx 10^(-3) cancel"kJ")#

#= -"133.44 atm/K"# where we multiplied by a ratio of universal gas constants to ensure the units worked out.

And if you look on the water phase diagram, this slope SHOULD be negative. This is because it requires **INPUT** of heat to melt (i.e. ** ice contracts when it melts**.

As a result, the left-hand side becomes:

#("2 atm")/(T_f' - "273.15 K") = -"133.44 atm/K"#

Solve for the **new freezing point** to get that:

#T_f' = "2 atm"/(-"133.44 atm/K") + "273.15 K"#

#=# #ul"273.14 K"#

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