Chemistry, 09.12.2020 23:10 vicada2782

How much rubidium would it take to equal the pressure wave of a stander hand grenade. (Don't ask, I need it for a project)

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Answer from: gacyber

Explanation:

rubidium is actually the pressure of the moons gravity times by monkey d. luffys meat capacity

Answer from: gabrielaperezcz

don't kill m

As a practical example we may consider an experiment involving the determination of the uptake of rubidium by portions of excised roots. A 1 × 10−3 M solution of RbCl2, i.e. each 1 ml of solution contains 1 µmole of rubidium, is labelled with 86Rb. If after mixing we take 1 ml of this solution and count it, we can quite simply determine its specific activity as x c.p.m./µmole Rb. (In practice for a 15–60 min experiment we should probably label the solution with about 50 µCi 86Rb/litre of uptake solution, i.e. about 0.05 µCi/ml, but note that it is quite immaterial to know the actual activity of the tracer in order to calculate the specific activity of the test solution). We may then allow 1 g weight samples of excised roots to absorb ions for a given unit time. After the elapse of the appropriate time each root sample is removed from solution and following appropriate procedures (see page 316 for typical details) the sample is counted, giving say y c.p.m. for 1 g roots. Then the amount of rubidium taken up by a root sample

(5)

Of course, this example has been simplified both in terms of procedure and solution concentration in order to bring out the essential principle, but such a basic calculation can be developed for a wide range of experimental conditions.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

16- counting by fours

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How much rubidium would it take to equal the pressure wave of a stander hand grenade. (Don't ask, I...