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sveticcg [70]
2 years ago
7

You are carefully watching the temperature of your melting point apparatus as it is heating up. At 132 C it is still a white sol

id. Your friend asks you a question, after answering you look back at the melting point apparatus and the temperature is 138 C and the solid is fully melted. What do you do?
Chemistry
1 answer:
Lisa [10]2 years ago
6 0

Answer:

See the answer below

Explanation:

<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>

1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.

2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.

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Which statement is TRUE regarding the macroscopic and
damaskus [11]

Answer:

Chemists make observations on the macroscopic a scale that lead to conclusions about microscopic features

Explanation:

Many important chemical observations are made on the macroscopic scale. This is because, many of the scientific equipments available are not presently able to provide direct evidence about microscopic processes. Evidences obtained from macroscopic observations could serve as important insights into the nature of certain microscopic processes.

This is evident in the study of the structure of the atom. Most of the evidences that led to the deduction of the atomic structure were obtained from macroscopic evidence but ultimately provided important information about the microscopic structure of the atom.

7 0
2 years ago
If the actual yield of the reaction was 75% instead of 100%, how many molecules of no would be present after the reaction was ov
artcher [175]
To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:

N₂ + O₂ → 2 NO

Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.

1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present

Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>
8 0
2 years ago
Read 2 more answers
A block of aluminum weighing 140 g is cooled from 98.4°C to 62.2°C with the release of 1080 joules of heat. From this data, calc
GrogVix [38]

<u>Answer:</u>

Specific heat of a substance is the value that describe how the added heat energy of substance has the impact on its temperature.

Unit is <em>(\frac {J}{Kg.K})</em>

<em>C = Q/m. ∆T</em>

<em>C – Specific heat (\frac {J}{Kg.K})</em>

<em>Q- heat energy (J)</em>

<em>M – Mass (Kg)</em>

<em>∆T- change in temperature (K) </em>

<u>Explanation:</u>

<em>Given data:</em>

<em>M= 140 g = 0.14 Kg</em>

<em>Q – 1080 Joules.</em>

<em>∆T – 98.4 – 62.2 = 36.2</em>

Substituting  the given data in Equation

<em>Specific heat of Aluminium  = \frac {1080}{(0.14 \times 36.2)} = 213.10 (\frac {J}{Kg.K})</em>

3 0
2 years ago
Read 2 more answers
How many moles of sodium bicarbonate is needed to neutralize 0.8 ml of sulphuric acid?
svet-max [94.6K]

Answer:

n NaHCO3 = 9.6 E-3 mol

Explanation:

balanced reaction:

  • 2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)
  • assuming a concentration of H2SO4 6M....normally worked in the lab

⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4

according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 ( sodium bicarbonate)

⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * ( 2 mol NaHCO3 / mol H2SO4 )

⇒ ,mol NaHCO3 = 9.6 E-3 mol

So 9.6 E-3 mol NaHCO3,  are the minimun moles necessary to neutralize the acid.

6 0
2 years ago
If a Lys residue that interacts with 2,3-bisphosphoglycerate (BPG) in the central cavity of hemoglobin is changed to a Ser resid
masha68 [24]

Answer: Option (c) is the correct answer.

Explanation:

When there will be interaction between 2,3-bisphosphoglycerate (BPG) and Lys residue in the central cavity of hemoglobin then it will change into Ser residue. After that, the it will lead to less stable formation of deoxygenated state (T state).

Since, serine is not a charged amino acid therefore, it will lead to a decrease in the binding affinity of hemoglobin for 2,3-bisphosphoglycerate.

As a result, stability of the T state will decrease.

Thus, we can conclude that in the given situation the T state would be less stable.

3 0
2 years ago
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