CHEMISTRY COLLEGE

Calculate the number of particles in 5.0 grams of NaCl.

Answers

Answer 1

Answer:

Answer:

5.2 × 10²² particles NaCl

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

5.0 g NaCl

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Na - 22.99 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

Step 3: Convert

$5.0 \ g \ NaCl((1 \ mol \ NaCl)/(58.44 \ g \ NaCl) )((6.022 \cdot 10^(23) \ particles \ NaCl)/(1 \ mol \ NaCl) )$ = 5.15229 × 10²² particles NaCl

Step 4: Check

We are given 2 sig figs. Follow sig fig rules and round.

5.15229 × 10²² particles NaCl ≈ 5.2 × 10²² particles NaCl

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10 POINTS : What are 4 things created from Big Bang?

Answers

Answer:

Explanation:

One second after the Big Bang, the universe was filled with neutrons, protons, electrons, anti-electrons, photons and neutrinos.Jun 17, 2017

4. Oxygenating myoglobin. The myoglobin content of some human muscles is about 8 g kg 1. In sperm whale, the myoglobin content of muscle is about 80 g kg 1 . (a) How much O 2 is bound to myoglobin in human muscle and in sperm whale muscle? Assume that the myoglobin is saturated with O 2, and that the molecular weights of human and sperm whale myoglobin are the same. Berg, Jeremy M.. Biochemistry (p. 213). W. H. Freeman. Kindle Edition.

Answers

Answer:

0.01454 grams of oxygen is present per kilogram of human muscle.

0.1454 grams of oxygen is present per kilogram of sperm whale muscle.

Explanation:

Given : The myoglobin is fully saturated with oxygen gas.

Moles of myoglobin = Moles of oxygen gas

Molecular mass of myoglobin = 17,600 g/mol

(assumed same fro whale and human )

Myoglobin content in humans = 8 g/kg

Mole of molyoglobin in human muscles per kg : $(8 g/kg)/(17,600 g/mol)$

Mass of oxygen present in per kg of human muscle:

Moles of oxygen gas × 32g/mol :

$(8 g/kg)/(17,600 g/mol)* 32 g/mol=0.01454$ g/kg of human muscle

Myoglobin content in whales= 80 g/kg

Mole of molyoglobin in whale muscles per kg : $(80 g/kg)/(17,600 g/mol)$

Moles of myoglobin = Moles of oxygen gas

Mass of oxygen present in per kg of sperm whale muscle:

Moles of oxygen gas × 32g/mol :

$(80 g/kg)/(17,600 g/mol)* 32 g/mol=0.1454$ g/kg of whale muscle

Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-chlorobutane. Based on this information, the major organic product(s) of this reaction are expected to be _____.

Answers

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

1,780 mL of a gas is at 37.5°C. At what temperature would the volume of the gasincrease to 2.55 L?

Answers

Answer:

171.8°C

Explanation:

V1= 1780ml, V2= 2.55L= 2550L, T1= 273+37.5= 310.5

T2=?

Applying

V1/T1 = V2/T2

1780/310.5 = 2550/T2

T2= 444.8K -273 = 171.8°C

VSEPR theory predicts that an atom with one lone pair and three bonding pairs (such as the N-atom in aniline) will have a tetrahedral electron geometry and a trigonal pyramidal molecular geometry due to steric repulsions between H-atoms and the N-atom lone pair. However, in question 5 you observed that the N-atom in aniline is not perfectly sp3 hybridized (i.e. the hybridization is different from that predicted for a tetrahedral electron geometry). Briefly describe all of the factors that result in the calculated hybridization of the N-atom lone pair

Answers

Answer: The lone pair of electron on nitrogen is accommodated in a 2p orbital hence it interacts with the pi system in aniline.

Explanation:

Aniline is less basic than amines. This is because, the nitrogen atom in aniline is not purely sp3 hybridized. Its actual hybrization state is closer to sp2 because the lone pair on nitrogen is accommodated in a 2p orbital. The nitrogen atim in aniline is planar and its

lonely pair interacts with the pi electron system of aniline. This makes the lone pair unavailable for protonation hence aniline is less basic than amines.

Final answer:

The calculated hybridization of the N-atom lone pair in aniline is affected by electron-electron repulsions, resonance, and steric effects from substituents on the aromatic ring.

Explanation:

The calculated hybridization of the N-atom lone pair in aniline is different from the predicted sp3 hybridization due to a combination of factors:

The presence of a lone pair on the nitrogen atom leads to electron-electron repulsions, causing distortions in the molecule's geometry.
The lone pair on the nitrogen atom can participate in resonance, resulting in delocalization of electrons and a change in hybridization.
The presence of the substituents on the aromatic ring can affect the hybridization of the N-atom lone pair by exerting steric effects.

Overall, these factors contribute to the observed hybridization of the N-atom lone pair in aniline, deviating from the predicted tetrahedral electron geometry.