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Evgen [1.6K]
3 years ago
13

The graph below shows how solubility changes with temperature

Chemistry
2 answers:
Svetradugi [14.3K]3 years ago
5 0

Answer:

Option C) Na2SO4 and Na2HAsO4

Explanation:

The solubility of salts can be affected by many things (pH, temperature, etc), this is why when comparing the solubility of two or more salts it's important to establish the conditions.

In this case the graph was made for changes in Temperature, so the first thing is to identify the T=40°C and then, which curves cross at that point.

As can be seen, the <u>red curve</u> (Na2HAsO4) and the <u>blue curve</u> (Na2SO4 ) cross at 40°C indicating that those compounds have similar solubilities at that temperature.

torisob [31]3 years ago
4 0
By looking at the graph, you can determine the answer is C. Na2HAsO4 and Na2SO4.
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In the rock cycle, where is the energy being released?
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Answer:

the two major sources of energy for the rock cycle are also shown; the sun provides energy for surface processes such as weathering, erosion, and transport, and the Earth's internal heat provides energy for processes like subduction, melting, and metamorphism.

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A 50.0g g sample of 16n decays to 12.5g in 14.4 seconds. What is its half life
mixas84 [53]
T is amount after time t 
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>

<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>

<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>

<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>

<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>

<span>HL = 14.4 s / 2 = 7.2 seconds </span>
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3 years ago
A belief judgement or way of thinking about something is called?
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Addition of water to an alkyne gives a keto‑enol tautomer product. Draw an enol that is in equilibrium with the given ketone
Marrrta [24]

Addition of water to an alkyne gives a keto‑enol tautomer product and that is the product changed into 2-pentanone, then the alkyne need to had been 1-pentyne. 2-pentyne might have given a combination of 2- and 3-pentanone.

<h3>What is the keto-enol means in tautomer?</h3>

They carries a carbonyl bond even as enol implies the presence of a double bond and a hydroxyl group. The keto-enol tautomerization equilibrium is depending on stabilization elements of each the keto tautomer and the enol tautomer.

  1. The enol that could provide 2-pentanone might had been pent-1- en - 2 -ol. Because an equilibrium favors the ketone so greatly, equilibrium isn't an excellent description.
  2. If the ketone have been handled with bromine, little response might be visible because the enol content material might be too low.
  3. If a catalyst have been delivered, NaOH for example, then formation of the enolate of pent-1-en - 2 - ol might shape and react with bromine.
  4. This might finally provide a bromoform product. Under acidic conditions, the enol might desire formation of the greater substituted enol constant with alkene stability.

7 0
2 years ago
cooling a sample of matter from 70°c to 10°c at constant pressure causes its volume to decrease from 873.6 to 712.6 cm3. classif
jek_recluse [69]

Explanation:

Expression for the coefficient of thermal expansion is as follows.

           \alpha = \frac{1}{V}(\frac{\Delta V}{\Delta T})

where,   V = initial volume

          \Delta V = Final volume - initial volume

                      = (712.6 - 873.6) cm^{3}

                      = -161 cm^{3}

Now, we will calculate the change in temperature as follows.

          \Delta T = Final temperature - Initial temperature

                       = (10 + 273) K - (70 + 273) K

                       = 283 K - 343 K

                       = -60 K

Substituting these values into the equation as follows.

     \alpha = \frac{1}{873.6} \times (\frac{161}{60}) K^{-1}

                 = 0.00307 K^{-1}

It is known that for non-ideal gases the value of alpha is 0.366% which is 0.00366 per Kelvin. As it is close to our result, hence the given sample of gas is a non-ideal gas.

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