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igomit [66]
3 years ago
9

In converting between units, it is never necessary to use more than one conversion factor. A. True B. False

Chemistry
2 answers:
lakkis [162]3 years ago
6 0
False, in converting between units, it is never necessary to use more than one conversion factor.
Radda [10]3 years ago
5 0

It is \boxed{{\text{true}}} that it is never necessary to use more than one conversion factor.

Further Explanation:

The measurements are compared with each other with the use of standards called units. These units are divided into two categories which are basic and derived units.

The units that do not require any other unit for expressing themselves and are used to define other units are known as basic units. Mass, time, length and temperature are some basic units. But some units require basic units to describe them and these are known as derived units. Density, volume, and area are some of the derived units.

Following are 7 basic quantities along with their units in the SI system:

1. Length: Meter (m)

2. Mass: Kilogram (kg)

3. Time: Second (s)

4. Thermodynamic temperature: Kelvin (K)

5. Electric current: Ampere (A)

6. Amount of substance: Mole (mol)

7. Luminous intensity: Candela (Cd)

One unit can be converted into another one with the help of a proper conversion factor. Conversion factors are the ratios that are written in the form of fractions and multiplication of which with the original unit gives the desired units.

Consider the conversion of 500 g of any substance into kg. The conversion factor for this conversion is,

 1{\text{ g}} = {10^{ - 3}}{\text{ kg}}

Therefore the desired conversion is done as follows:

\begin{aligned}{\text{Mass of object}} &= \left( {500{\text{ g}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ kg}}}}{{1{\text{ g}}}}} \right)\\&= 0.5{\text{ kg}}\\\end{aligned}

This is a single-step conversion. So it is never necessary to use more than one conversion factor.

Learn more:

  1. What is the mass of 1 mole of viruses: brainly.com/question/8353774
  2. Determine the moles of water produced: brainly.com/question/1405182

Answer details:

Grade: Senior School

Chapter: Basic concepts of chemistry

Subject: Chemistry

Keywords: unit, basic, fundamental, mass, time, thermodynamic temperature, luminous intensity, amount of substance, electric current, conversion factor.

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An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm
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The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

 xA= 0.34, yA= 0.55

ii)      0.50 = 0.55 nv + 0.34 nL

     Therefore, nV =    0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

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Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

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Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

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What is the energy change associated with 1.5 mole of D being formed? 
AleksandrR [38]

Answer:

–36 KJ.

Explanation:

The equation for the reaction is given below:

2B + C —› D + E. ΔH = – 24 KJ

From the equation above,

1 mole of D required – 24 KJ of energy.

Now, we shall determine the energy change associated with 1.5 moles of D.

This can be obtained as illustrated below:

From the equation above,

1 mole of D required – 24 KJ of energy

Therefore,

1.5 moles of D will require = 1.5 × – 24 = –36 KJ.

Therefore, –36 KJ of energy is associated with 1.5 moles of D.

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