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anygoal [31]
4 years ago
11

A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu

re of the object was –3.5 ºK. Identify two errors in the student’s recorded temperature.
Chemistry
1 answer:
bogdanovich [222]4 years ago
6 0

Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

Explanation:

The temperature of an object can be written using different temperature scales.

The two most important scales are:

- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).

- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

T(K)=T(^{\circ}C)+273.15

Therefore  in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

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Benzene is often produced as a side product during grignard reactions using phenylmagnesium bromide. how can its formation be ex
Alex_Xolod [135]

Benzene is produced as a side product when water react with phenyl magnesium bromide during Grignard reactions.

<h3>What is Grignard Reaction ? </h3>

The Grignard reaction  involves an R-Mg-X, a carbon chain bound to a magnesium halide, used to form alcohols  by attacking carbonyls such as in aldehydes or ketones.

Here R act as a nucleophile and Mg-X act as a electrophile. Instead of attacking carbonyl, which is  present in the reaction, the benzene ring will attack  water molecules present around, grabbing a H⁺ and pushing away MgBr .    

The required reaction is :

PhMgBr + H₂O → Ph-H + MgBrOH

Thus from the above conclusion we can say that it is always advised to use dry Grignard to avoid this reaction.

Learn more about the GRIGNARD REACTION here :

brainly.com/question/23971610

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4 0
1 year ago
Calculate the amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C.
qwelly [4]

Answer:

The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

Between heat and temperature there is a direct proportional relationship. The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

  • c= 4.184 \frac{J}{g*C}
  • m= 32 g
  • ΔT= Tfinal - Tinitial= 22°C - 8°C= 14°C

Replacing:

Q= 32 g* 4.184 \frac{J}{g*C} *14 °C

Solving:

Q= 1,874.432 J

<u><em>The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J</em></u>

7 0
3 years ago
How many electrons are in the 4p orbital does Br possesses?
Klio2033 [76]

Answer:

5 electron

It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital.

5 0
4 years ago
Identify the products formed in this Brønsted-Lowry reaction. <br> HCO−3+BrO−↽−−⇀acid+base
olganol [36]

Answer:

HCO3- + BrO- → HBrO + CO32-

Explanation:

I think this is the answer hope it helps

8 0
3 years ago
Calculate the pH of a 0.10 M solution of acidic acid HC2H3O2(aq) at 25 °C. Ka for HC2H3O2 = 1.8×10−5 at 25 °C g
TiliK225 [7]

Answer:

The answer is "3".

Explanation:

acetic acid Dissociation:  

CH_3COOH \ < - - - - - - - - - - - - - \ > CH_3COO^{-}+H^{+}\\\\

Dissociation constant of the Ka:

Ka = \frac{[CH3COO^{-}] [H^{+}]}{[CH_3COOH]}\\\\

using the ICE table:

CH_3COOH \ < - - - - - - - - - - - - - \ > CH_3COO^{-}+H^{+}\\\\ \to 1.8 \times 10^{-5} =\frac{( x  \times x) }{0.1-x} -------(2)

x is negligible compared to Ka.

1.8\times 10^{-5} = \frac{x^2}{0.10}\\\\x^2 = 1.8 \times 10^{-6}\\\\x = 1.34 \times 10^{-3}\\\\pH = -\log [H^{+}]

From the ICE table, [H^{+}] = x = 1.34 \times 10^{-3}

\to pH = -\log(1.34 \times 10^{-3}) = 3

5 0
4 years ago
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