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klasskru [66]
3 years ago
7

Ammonium chloride is used as an expectorant in cough medicine. It has a density of 1.53 g/cm^3. What is the mass of 26.0L of thi

s substance in Kg
Chemistry
1 answer:
Katena32 [7]3 years ago
8 0

Answer:

                     39.78 Kg

Explanation:

Data Given;

                  Density  =  1.53 g/cm³

                  Mass  =  <u>??</u>

                  Volume  =  26.0 L  =  26000 cm³

Formula Used;

                       The measure of mass per unit volume is called as density. It is a physical property of a substance and tells how tightly the particles are packed. Mathematically;

                  Density  =  Mass ÷ Volume

Solving for Mass,

                  Mass  =  Volume × Density

Putting values,

                  Mass  =  26000 cm³ × 1.53 g.mL⁻¹

                  Mass  =  39780 g  or   39.78 Kg

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Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)
Sedbober [7]

Answer:

The yield of the product in gram is \mathsf{{w_P}=0.26 \ gram}

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = 2.7 \times 10^{-3} \ moles

We know that:

Number of moles = mass/molar mass

Then:

2.7 \times 10^{-3} = \dfrac{ mass}{257.997}

mass = 2.7 \times 10^{-3} \times 257.997

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = \dfrac{n_P}{n_R}\times 100\%

where;

n_P = \dfrac{w_P}{m_P}    and n_R = \dfrac{w_R}{m_R}

Theoretical yield = \dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

Given that the theoretical yield = 100%

Then:

100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}

{w_P}=\dfrac{w_R \times m_P}{m_R}

where,

w_P = derived weight of the product

m_P =the molecular mass of the derived product

m_R = the molecular mass of the reagent

w_R = weight in a gram of the reagent

{w_P}=\dfrac{w_R \times m_P}{m_R}

{w_P}=\dfrac{0.697 \times 96.2}{257.997}

\mathsf{{w_P}=0.26 \ gram}

8 0
3 years ago
Luca wants to identify and quantity the chemicals causing air pollution in a city, but he is using a manual air sampler to colle
xenn [34]

Answer

Manual samplers are prone to effects of temperature, speed of wind and air concentrations.

Explanation

Manual samplers face several challenges that can act as drawbacks to obtaining accurate results. They are subjected to effects of sampling duration where long sampling times are needed to obtain adequate mass for detection. Manual samplers face challenges when measuring non-volatile species because particles are observed into the adsorption medium at a slower rate of diffusion.

5 0
3 years ago
Explain how heat would change the density of a parcel of air
BartSMP [9]
<span>Heat would make the molecules move faster so they would spread out therefore making air less dense.
Less dense means less packed. Heat increase the kinetic energy</span>
6 0
3 years ago
What is the molarity of a solution containing 501mL with 35g NaCl
Vesna [10]

Answer:

Data:

mass of solute: 35g of NaCl

m.mass of solute: 58g/mol

volume of solution: 501mL

Molarity=?

Explanation:

501ml = 0.5dm3

M= g of solute/m.mass of solute*vol of solution

M= 35/58*0.5

M=1.20

6 0
3 years ago
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insens350 [35]
Reliable results are results that can be... A: Communicated
4 0
3 years ago
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