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2 years ago

Pre-lab Assignment

Chemistry

1 answer:

alina1380 [7]2 years ago

3 0

Answer:

1. d. Person A rounded their measurement

2) a. 105.

b. 76.4

c. 0.04

d. 1.988

Explanation:

1. Measuring an object is done with the least precise one. Person A recorded 1.5cm while Person B recorded 1.50cm of the same object. This measurement differs in the sense that Person A rounded up their measurement to 2 significant figures (1.5) while Person B recorded their answer as three significant figures (1.50).

2. During multiplication and division, the result is written as the measurement in the lowest significant figure.

a. 7.00 x 15.00 = 105. Calculator showed 105.00, which is 5 s.f (two zeros after decimal point are significant). The least significant figure is 3, hence 105.00 is rounded up to 105.

b. 40.2.1.901 = 76.4; The calculator answer of this operation is 76.420200 but since the lowest significant figure is 3. We round up our answer to 3 s.f i.e. 76.4

c. (0.003)(13.2) = 0.04. For this operation, calculator answer is 0.0396, however, the lowest significant figure in the question is 1 i.e. 0.003. Note that the two zeros before 3 are insignificant. Hence, our answer will be rounded up to 1 s.f i.e. 0.04

d. 1.590+ 0.3975 = 1.988. the calculator answer to this operation is 1.9875. However, the least significant figure in the question is 4 i.e. 1.590. hence, the answer will be rounded up to 4 s.f which is 1.988.

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An unknown compound melting at 131 - 133 C. It is thought to be one of the following compounds: trans-cinnamic acid (133-134); b

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Answer:

benzamide

Explanation:

Compound melting Point ,ºC Melting Pont Mixture, ºC

X 131 - 133

trans-cinnamic 133 - 134 110 - 120

acid

benzamide 128 - 130 130-132

malic acid 131 -133 114 -124

Benzoin 135 - 137 108 - 116

The compound X is benzamide since the melting point range is the one closest to this compound ( 130-132 ºC)

The reason there is not an exact match is not due due to the presence of impurities. The presence of impurities always lower the melting point ( it is a coligative property such as the melting point depresion of salt and water )

The reason for the deviation must be be some other factors such as preparation of the sample in the capillary, errors in reading the thermometer, rate of heating, etc.

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2 years ago

Explain one way the water cycle affects climate. Use complete sentences.

ladessa [460]

Answer:

The water cycle will affect the climate by causing rain and snow.

Explanation:

if tempatures increase more rain fall will occur because more water evaporates in the air.

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1 year ago

Calculate the daily aluminum production of a 150,000 [A] aluminum cell that operates at a faradaic efficiency of 89%. The cell r

Gala2k [10]

Explanation:

It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.

$24 \times 3600 sec$

Hence, total charge passed daily is calculated as follows.

$150,000 \times 24 \times 3600 sec$

And, number of Faraday of charge is as follows.

$\frac{150,000 \times 24 \times 3600 sec}{96500}$

= 134300.52 F

The oxidation state of aluminium in $Al_{2}O_{3}$ is +3.

$Al^{3+} + 3e^{-} \rightarrow Al(s)$

So, if we have to produce 1 mole of Al(s) we need 3 Faraday of charge.

Therefore, from 134300.52 F the moles of Al obtained with 89% efficiency is calculated as follows.

$\frac{134300.52 F}{3} \times \frac{89}{100}$

= 39842.487 mol

or, = $3.9842 \times 10^{4} mol$

Molar mass of Al = 27 g/mol

Therefore, mass in gram will be calculated as follows.

Mass in grams = $3.9842 \times 10^{4} mol \times 27$

= $107.57 \times 10^{4} g$

= 1075.7 kg/day

Thus, we can conclude that the daily aluminum production of given aluminium is 1075.7 kg/day.

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2 years ago

Consider the solutions, 0.04 m urea [(NH2)2C=O)], 0.04 m AgNO3 and 0.04 m CaCl2. Which has (i) the highest osmotic pressure, (ii

Lana71 [14]

Answer:

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

Explanation:

The colligative properties depend upon the number of solute particles in a solution.

The following four are the colligative properties:

a) osmotic pressure : more the concentration of the solute, more the osmotic pressure

b) vapor pressure: more the concentration of the solute, lesser the vapor pressure.

c) elevation in boiling point: more the concentration of the solute, more the boiling point.

d) depression in freezing point: more the concentration of the solute, lesser the freezing point.

the number of particle produced by urea = 1

the number of particle produced by AgNO3 = 2

the number of particle produced by CaCl2 = 3

As concentrations are same, CaCl2 will have more number of solute particles and urea will have least

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

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Answer:

Explanation:

Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.

In the living E. coli cells,

[ATP] = 7.9 mM;

[ADP] = 1.04 mM,

[glucose] = 2 mM,

[glucose 6-phosphate] = 1 mM.

Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

The reaction is given as

Glucose + ATP → glucose 6-phosphate + ADP

Now reaction quotient for given equation above is

$q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}$

$q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}$

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$q$ ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq

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