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mars1129 [50]
3 years ago
9

Measure the diameter of the circle using Ruler A and Ruler B.Given that the actual diameter of the circle is 2.264 cmcm, classif

y the following statements that describe the diameter measurement of the circle according to the ruler. Drag the appropriate items to their respective bins.
Chemistry
1 answer:
leva [86]3 years ago
5 0

Answer:

<u>Ruler A :</u>

According to ruler A, the diameter of the circle has only one certain digit and one uncertain digit. if we look at measurement of a diameter, it contain two significant figures. so the certainty  of the diameter measurement is <em>smaller. </em>

<u>Ruler B :</u>

According to ruler B, the certainty  of the diameter measurement is <em>greater. </em>because it contain two certain and one uncertain digit. it has three significant figures.

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FIRST ANSWER = BRAINLIEST
STALIN [3.7K]

A_{r} = 24.3

The average atomic mass of X is the <em>weighted average</em> of the atomic masses of its isotopes.  

We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its % abundance).  

Thus,  

0.790 × 24 u = 18.96 u

0.100 × 25 u =   2.50 u

0.110 × 26 u =    <u>2.86 u</u>  

       TOTAL =  24.3   u

∴ The relative atomic mass of X is 24.3.


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3 years ago
In a solution of a carbonated beverage the water is what? Solute, saturated, solvent or precipitate
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Https://us-static.z-dn.net/files/d49/33d4ec86853ef95e6f6c14242c663be4.png
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I need help!!! Calculating pH, pOH, [H+] and [OH-]
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Find the hydroxide ion concentration of a solution with a pOH of 5.90. To solve this, use a scientific calculator and enter 5.90 and use the +/- button to make it negative and then press the 10x key.
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3 years ago
At what temperature is the following reaction feasible: HCl(g) + NH3(g) -&gt; NH4Cl(s)?
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Energy is distributed not just in translational KE, but also in rotation, vibration and also distributed in electronic energy levels (if input great enough, bond breaks).

All four forms of energy are quantised and the quanta ‘gap’ differences increases from trans. KE ==> electronic.

Entropy (S) and energy distribution: The energy is distributed amongst the energy levels in the particles to maximise their entropy.

Entropy is a measure of both the way the particles are arranged AND the ways the quanta of energy can be arranged.

We can apply ΔSθsys/surr/tot ideas to chemical changes to test feasibility of a reaction:

ΔSθtot = ΔSθsys +  ΔSθsurr

ΔSθtot must be >=0 for a chemical change to be feasible.

For example: CaCO3(s) ==> CaO(s) + CO2(g) 

ΔSθsys = ΣSθproducts – ΣSθreactants 

ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) 

ΔSθsurr is –ΔHθ/T(K) and ΔH is very endothermic (very +ve),

Now ΔSθsys is approximately constant with temperature and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall.

But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800oCΔSθtot becomes plus overall (and ΔGθ becomes negative), so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.

CaCO3(s) ==> CaO(s) + CO2(g)  ΔHθ = +179 kJ mol–1  (very endothermic)

This important industrial reaction for converting limestone (calcium carbonate) to lime (calcium oxide) has to be performed at high temperatures in a specially designed limekiln – which these days, basically consists of a huge rotating angled ceramic lined steel tube in which a mixture of limestone plus coal/coke/oil/gas? is fed in at one end and lime collected at the lower end. The mixture is ignited and excess air blasted through to burn the coal/coke and maintain a high operating temperature.
ΔSθsys = ΣSθproducts – ΣSθreactants
ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) = (40.0) + (214.0) – (92.9) = +161.0 J mol–1 K–1
ΔSθsurr is –ΔHθ/T = –(179000/T)
ΔSθtot = ΔSθsys +  ΔSθsurr
ΔSθtot = (+161) + (–179000/T) = 161 – 179000/T
If we then substitute various values of T (in Kelvin) you can calculate when the reaction becomes feasible.
For T = 298K (room temperature)

ΔSθtot = 161 – 179000/298 = –439.7 J mol–1 K–1, no good, negative entropy change

For T = 500K (fairly high temperature for an industrial process)

ΔSθtot = 161 – 179000/500 = –197.0, still no good

For T = 1200K (limekiln temperature)

ΔSθtot = 161 – 179000/1200 = +11.8 J mol–1 K–1, definitely feasible, overall positive entropy change

Now assuming ΔSθsys is approximately constant with temperature change and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall. But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800–900oC ΔSθtot becomes plus overall, so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.
You can approach the problem in another more efficient way by solving the total entropy expression for T at the point when the total entropy change is zero. At this point calcium carbonate, calcium oxide and carbon dioxide are at equilibrium.
ΔSθtot–equilib = 0 = 161 – 179000/T, 179000/T = 161, T = 179000/161 = 1112 K

This means that 1112 K is the minimum temperature to get an economic yield. Well at first sight anyway. In fact because the carbon dioxide is swept away in the flue gases so an equilibrium is never truly attained so limestone continues to decompose even at lower temperatures.

8 0
3 years ago
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3 years ago
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