Chemistry, 03.11.2020 19:50 aurora32

Please help thank you!!

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Answer from: xrivas

None of the above

Answer from: dbhuggybearow6jng

I believe its B

Explanation:

Hope this helps!

-BB

Answer from: Quest

a mercury barometer is a device that uses the physical properties of mercury in a tube to measure barometric pressure. mercury barometers are used for : -

1. measurement of atmospheric pressure

2. determining the altitude of a place

3. weather forecasting

4. calibration and checking of aneroid barometers

5. application in fluid mechanics

6. measurement of pressure in aircrafts

7. surface weather analysis

8. preparation of barographs

9. usage in physical sciences like physics, astronomy & chemistry

10. preparation of aircraft altimeters

characteristics of mercury barometer : -

1.used to define and measure atmospheric pressure.

2.on the average at sea level the column of mercury rises to a height of about 760 mm.

3.this quantity is equal to 1 atmosphere.

4.it is also known as standard atmospheric pressure.

5.a barometer in which the weight of a column of mercury in a glass tube with a sealed top is balanced against that of the atmosphere pressing on an exposed cistern of mercury at the base of the mercury column, the height of the column varying with atmospheric pressure.

Write 15 uses of a mercury barometer along with its characterestics

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 :

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

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