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## What is the tool that scientists commonly use to measure liquid volume?

Truong-Son N.
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 $100.00 \pm \text{0.02 mL}$, when you properly measure to the index mark.)

• 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 $10.00 \pm \text{0.01 mL}$ 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.

## When oxygen combines with any alkali metal, M, what is the formula of the compound produced usually?

Nathan L.
Featured 5 months ago

The basic formula is color(red)("M"_2"O".

#### Explanation:

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 color(red)(6 valence electrons:

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 $8$ electrons in its outermost shell.

For an oxygen atom, the easiest way to get $8$ electrons in the outer shell is to obtain $2$ more electrons. This will in turn cause the sfcolor(blue)("oxygen" to have an ionic charge of color(blue)(2-, because there are color(blue)(2 more electrons than protons now.

$- - - - -$

For the alkali metals, which are located in group 1 of the periodic table, they each have $1$ valence electron. Here's the electrons-by-shell for potassium, one of the alkali metals:

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 color(red)("M" for the alkali metal) to have an ionic charge of color(red)(1+, because there is color(red)(1 less electron than the number of protons.

The formula:

The compound the alkali metal color(red)("M" and oxygen color(blue)("O" form must be electrically neutral. That is to say, the charges must balance out to $0$.

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 ${\text{O}}^{2 -}$), $2 -$ as the subscript of the cation, and the charge of the cation (the positive ion, in this case ${\text{M}}^{+}$) as the subscript of the *anion*. Here's what I mean, using sodium ($\text{Na}$) as an example:

sfcolor(green)("Therefore, the formula for any alkali metal M and oxygen (O)"
sfcolor(green)(" is M"_2"O".

## A moon rock collected by a U.S. Apollo mission is estimated to be 4.20 billion years old by uranium/lead dating. Assuming that the rock did not contain any lead when it was formed, what is the current mass of 206Pb per 1.235 g of 238U in the rock?

Michael
Featured 4 months ago

0.980 g

#### Explanation:

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 $\alpha$ and 6 $\beta$ decays:

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:

$\textsf{{U}_{t} = {U}_{0} {e}^{- \lambda \text{t}}}$

$\textsf{{U}_{0}}$ is the number of undecayed U 238 atoms initially.

$\textsf{{U}_{t}}$ is the number of undecayed U 238 after time $\textsf{t}$.

$\textsf{\lambda}$ is the decay constant.

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

$\textsf{{U}_{0} = {U}_{t} + P {b}_{t}}$

Where $\textsf{P {b}_{t}}$ is the number of Pb 206 atoms formed after time $\textsf{t}$.

The decay equation can therefore be written:

$\textsf{{U}_{t} = \left({U}_{t} + P {b}_{t}\right) {e}^{- \lambda \text{t}}}$

$\therefore$$\textsf{\frac{{U}_{t}}{\left({U}_{t} + P {b}_{t}\right)} = {e}^{- \lambda \text{t}}}$

$\therefore$sf((U_(t)+Pb_(t))/(U_(t))=e^(lambdat)

$\therefore$$\textsf{1 + \frac{P {b}_{t}}{U} _ \left(t\right) = {e}^{\lambda \text{t}}}$

$\therefore$$\textsf{\frac{P {b}_{t}}{U} _ \left(t\right) = {e}^{\lambda \text{t}} - 1}$

The half - life, $\textsf{{t}_{\frac{1}{2}}}$, of U 238 is $\textsf{4.468 \times {10}^{9}}$ years.

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

sf(lambda=0.693/(t_(1/2))

$\textsf{\lambda = \frac{0.693}{4.468 \times {10}^{9}} = 0.1551 \times {10}^{- 9} {\text{ " "yr}}^{- 1}}$

We can get the number of moles of $\textsf{n}$ each isotope by dividing the mass by the mass of one mole, which is the $\textsf{{A}_{r}}$ in grams:

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

$\textsf{{n}_{P {b}_{t}} = \frac{m}{205.974449}}$

$\textsf{{n}_{{U}_{t}} = \frac{1.235}{238.050784} = 0.0051879}$

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.

$\therefore$$\textsf{\frac{\frac{m}{205.974449}}{0.0051879} = {e}^{\lambda \text{t}} - 1}$

$\therefore$$\textsf{\frac{m}{1.068589} = {e}^{\lambda t} - 1}$

$\therefore$$\textsf{\frac{m}{1.068599} + 1 = {e}^{\lambda t}}$

Taking natural logs of both sides $\Rightarrow$

$\textsf{\ln \left(\frac{m}{1.068599} + 1\right) = \lambda t}$

Putting in the values for $\textsf{\lambda}$ and $\textsf{t \Rightarrow}$

$\textsf{\ln \left(\frac{m}{1.068599} + 1\right) = 0.1551 \times {10}^{- 9} \times 4.2 \times {10}^{9} = 0.651}$

$\therefore$$\textsf{\frac{m}{1.068599} + 1 = 1.91746}$

$\therefore$$\textsf{\frac{m}{1.068599} = 0.91746}$

$\textsf{m = 0.91746 \times 1.068599 \textcolor{w h i t e}{x} g}$

$\textsf{m = 0.980 \textcolor{w h i t e}{x} g}$

## What information does the chemical equation 2C_4 H _10+13O_2 -> 8CO_2 + 10H_2 O provide?

Doc048
Featured 3 months ago

see explanation => ...

#### Explanation:

The following is for posted equation ...

## A first order reaction has a half-life of 43.80 s. How long will it take for until 10% of the reactant remains?

Truong-Son N.
Featured 1 month ago

$t = \text{145.5 s}$

(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

${t}_{\text{1/2" = "43.80 s}}$,

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

• After one half-lives, $\left[A\right] = \frac{1}{2} {\left[A\right]}_{0}$.
• After two half-lives, $\left[A\right] = \frac{1}{4} {\left[A\right]}_{0}$.
• After three half-lives, $\left[A\right] = \frac{1}{8} {\left[A\right]}_{0}$.
• After four half-lives, $\left[A\right] = \frac{1}{16} {\left[A\right]}_{0}$.

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

$\left[A\right] = {\left(\frac{1}{2}\right)}^{n} {\left[A\right]}_{0}$

where $n$ is the number of half-lives that passed.

The number of half-lives passed, $n$, is simply the total time divided by the known half-life.

Thus, $n = t / {t}_{\text{1/2}}$, and we have the current concentration as a function of the starting concentration, half-life, and total time passed:

$\overline{\underline{|}} \stackrel{\text{ ")(" "[A] = (1/2)^(t//t_"1/2") [A]_0" }}{|}$

Given that 10% of the sample remains, we know that

$\left[A\right] = 0.1 {\left[A\right]}_{0}$,

and thus,

${\left(\frac{1}{2}\right)}^{t / {t}_{\text{1/2}}} = 0.1$.

To solve for $t$, take the $\ln$ of both sides.

$\ln {\left(\frac{1}{2}\right)}^{t / {t}_{\text{1//2}}} = \ln 0.1$

$\frac{t}{{t}_{\text{1/2}}} \ln \left(\frac{1}{2}\right) = \ln 0.1$

Therefore, it will take this long for 10% of the sample to remain:

$\textcolor{b l u e}{t} = {t}_{\text{1/2}} \cdot \ln \frac{0.1}{\ln} \left(\frac{1}{2}\right)$

$= \text{43.80 s} \cdot \frac{\ln 0.1}{\ln \left(\frac{1}{2}\right)}$

$=$ $\textcolor{b l u e}{\text{145.5 s}}$

METHOD 2

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

${t}_{\text{1/2}} = \frac{\ln 2}{k}$

And thus, the rate constant is given by:

$k = \frac{\ln 2}{t} _ \text{1/2}$

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

$\ln \left[A\right] = - k t + \ln {\left[A\right]}_{0}$,

one can then solve for the time passed simply by knowing that $\left[A\right] = 0.1 {\left[A\right]}_{0}$ as mentioned earlier.

$\ln \left(0.1 {\left[A\right]}_{0}\right) - \ln {\left[A\right]}_{0} = - k t$

$\ln \left(\frac{0.1 \cancel{{\left[A\right]}_{0}}}{\cancel{{\left[A\right]}_{0}}}\right) = - k t$

$t = \ln \frac{0.1}{- k}$

$= \frac{\ln 0.1}{- \frac{\ln 2}{{t}_{\text{1/2}}}}$

$= {t}_{\text{1/2}} \cdot \frac{\ln 0.1}{\ln \left(\frac{1}{2}\right)}$

which is of the same form as in method 1. We would also get $\text{145.5 s}$ in this way.

## Calculate the change in melting point of ice at 0^@ "C" when pressure is increased by "2 atm"?

Truong-Son N.
Featured 1 month ago

$\Delta {T}_{f} \approx - {0.015}^{\circ} \text{C}$

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 ${10}^{10} \text{ergs}$ in $\text{1 kJ}$, so you should have had $\Delta {H}_{\text{fus" = 796 * 4.186 * 10^6 "erg/g}}$, or $\text{6.02 kJ/mol}$.
• 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 ${\text{1 g/cm}}^{3}$... and they are:

${\rho}_{w} = \frac{1}{1.0001} {\text{g/cm"^3 = "0.9999 g/cm}}^{3}$

${\rho}_{\text{ice" = 1/1.0908 "g/cm"^3 = "0.9168 g/cm}}^{3}$

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

$\frac{\mathrm{dP}}{\mathrm{dT}} = \left(\Delta {\overline{H}}_{\text{fus")/(T_fDeltabarV_"fus}}\right)$

where:

• $\frac{\mathrm{dP}}{\mathrm{dT}}$ is the slope of the two-phase coexistence curve on a phase diagram.
• $\Delta {\overline{H}}_{\text{fus}}$ is the change in molar enthalpy of fusion in $\text{kJ/mol}$.
• ${T}_{f}$ is the freezing point of the substance in $\text{K}$.
• $\Delta {\overline{V}}_{\text{fus" = barV_w - barV_"ice}}$ is the change in molar volume in $\text{L/mol}$ due to melting ice.

We already converted $\Delta {\overline{H}}_{\text{fus}}$ to the proper units. Now for the densities.

${\overline{V}}_{w} = {\left[\left(0.9999 \cancel{\text{g H"_2"O"))/cancel("cm"^3) xx cancel("1 cm"^3)/cancel"mL" xx (1000 cancel"mL")/"L" xx ("1 mol")/(18.015 cancel("g H"_2"O}}\right)\right]}^{- 1}$

= 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 $\frac{\mathrm{dP}}{\mathrm{dT}}$.

In some small interval where the temperature and pressure change, we consider the reference temperature and pressure to be ${0}^{\circ} \text{C}$ at $\text{1 atm}$, the normal melting point:

$\frac{\mathrm{dP}}{{\mathrm{dT}}_{f}} \approx \frac{\Delta P}{\Delta {T}_{f}} = \left(\text{3 atm" - "1 atm")/(T_f' - "273.15 K}\right)$

The right-hand side of the equation becomes:

$\frac{\Delta P}{\Delta {T}_{f}} \approx \left(6.02 \cancel{\text{kJ""/"cancel"mol")/("273.15 K" cdot (0.01802 cancel"L""/"cancel"mol" - 0.01965 cancel"L""/"cancel"mol")) xx (0.082057 cancel"L"cdot"atm")/(8.314472 xx 10^(-3) cancel"kJ}}\right)$

$= - \text{133.44 atm/K}$

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. $\Delta {\overline{H}}_{\text{fus}} > 0$, while $\Delta {\overline{V}}_{\text{fus}} < 0$ and $T > 0$. This would be experimentally showing that 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:

${T}_{f} ' = \text{2 atm"/(-"133.44 atm/K") + "273.15 K}$

$=$ $\underline{\text{273.14 K}}$

Or $\textcolor{b l u e}{\Delta {T}_{f} \approx - {0.015}^{\circ} \text{C}}$. It should make sense that the increase in pressure makes ice easier to melt; the added pressure is like you squeezing hard on the ice.

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