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nordsb [41]
3 years ago
13

A student sets up the following equation to convert a measurement.

Chemistry
1 answer:
Sholpan [36]3 years ago
6 0

Answer:

–0.13 Pa.m²

Explanation:

From the question given above, the following data were obtained:

Measurement (Pa.mm²) = –1.3×10⁵ Pa.mm²

Measurement (Pa.m²) =?

We can convert from Pa.mm² to Pa.m² by doing the following:

1 Pa.mm² = 1×10¯⁶ Pa.m²

Therefore,

–1.3×10⁵ Pa.mm² = –1.3×10⁵ Pa.mm² × 1×10¯⁶ Pa.m² / 1 Pa.mm²

–1.3×10⁵ Pa.mm² = –0.13 Pa.m²

Thus, –1.3×10⁵ Pa.mm² is equivalent to –0.13 Pa.m².

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Predict the sign of the entropy change of the system for each of the following reactions.
storchak [24]

Answer :

(a) The entropy change will also decreases.

(b) The entropy change will also increases.

(c) The entropy change will also decreases.

(d) The entropy change will also increases.

Explanation :

Entropy : It is defined as the measurement of randomness or disorderedness in a system.

The order of entropy will be,

As we are moving from solid state to liquid state to gaseous state, the entropy will be increases due to the increase in the disorderedness.

As we are moving from gaseous state to liquid state to solid state, the entropy will be decreases due to the decrease in the disorderedness.

(a) N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

In this reaction, the randomness of reactant molecules are more and as we move towards the formation of product the randomness become less that means the degree of disorderedness decreases. So, the entropy change will also decreases.

(b) CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

In this reaction, the randomness of reactant molecules are less and as we move towards the formation of product the randomness become more that means the degree of disorderedness increase. So, the entropy change will also increases.

(c) 3C_2H_2(g)\rightarrow C_6H_6(g)

In this reaction, 3 mole of gaseous C_2H_2 react to give 1 mole of gaseous C_6H_6 that means randomness become less that means the degree of disorderedness decreases. So, the entropy change will also decreases.

(d) Al_2O_3(s)+3H_2(g)\rightarrow 2Al(s)+3H_2O(g)

In this reaction, the number of moles of gases are same on both side of the reaction but 1 mole of solid Al_2O_3 react to give 2 moles of solid aluminium that means randomness become more that means the degree of disorderedness increases. So, the entropy change will also increases.

5 0
3 years ago
5. Calculate the decrease in temperature when 6.00 L at 20.0 °C is compressed to 4.00 L.
Kazeer [188]

Answer: 195.3 K

Explanation: Use Charles Law to get T2

Charles Law V1 / T1 = V2 / T2

Derive the equation to find T2

T2 = V2T1 / V1

Convert temperature into K

T1 = 20.0 °C + 273 = 293 K

Substitute the values:

4.00 L x 293 K / 6.00 L = 195. 3 K

5 0
3 years ago
How does a purebred individual differ from a hybrid individual?
zimovet [89]
Im gonna go based on the content of their blood. So as the generation continues, a purebred's blood has not differentiated too much and contains 98-100% of a "pure" bloodline meaning no outsiders of where this purebred originated has mixed. A hybrid is where the original bloodline is mixed with many others, so while a purebred has 99% of the original bloodline and the hybrid has 50% and he other half belongs to an entirely different line.
5 0
3 years ago
A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate for
laiz [17]

Answer:

CaF2 will not precipitate

Explanation:

Given

Volume of Ca(NO3)2 = 10 ml

Molar concentration of Ca(NO3)2 = 0.001

Volume of NaF = 10 ml

Molar concentration of  NaF  = 0.0001

Ksp for CaF2 = 3.2 * 10^ {-11}

CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2

Moles of calcium ion

= 10 * 0.001\\= 0.01

[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}

Moles of F- ion

= 10 * 0.0001\\= 0.001

[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}

Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}

Q is lesser than Ksp value of CaF2. Hence it will not precipitate

5 0
3 years ago
What property of liquid oxygen makes it especially difficult and potentially harmful to work with at home?
irga5000 [103]
The property of liquid oxygen that makes it especially difficult and potentially harmful to work with at home would be its cryogenic temperature. Liquid oxygen is being produced from the compression of oxygen gas to -196 degrees Celsius. As you can see, it has a very cold temperature that is why it used in cryogenics. Although liquid oxygen is non-toxic to humans, it would cause burns that are severe when being touched. Also, it would make certain materials brittle and unstable. Another property that makes it dangerous for use at home would be that it is very flammable. Proper handling is a must for this substance.
3 0
3 years ago
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