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3 years ago

Compare a base unit and a derived unit, and list the derived units used for density and volume.

Chemistry

1 answer:

KIM [24]3 years ago

4 0

Answer:

Explanation:

The fundamental units of a measurement is known as its base unit. The units of these substances serves as the base through which other quantities depends. Examples of such quantities are mass, length, time, electric current, temperature, amount of substance and luminous intensity.

Derived units are those that results from the combination of the fundamental or basic units. Examples of derived quantities are force, volume, density, pressure e.t.c.

Derived units of Density:

kgm⁻³

gcm⁻³

Derived unit of volume:

m³

cm³

dm³

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How much carbon would be needed to make 8.8g of carbon dioxide?<br>

PIT_PIT [208]

The calculation for such a question can be achieved via Avogadro hypothesis

We know molar mass of CO2 is 44g/mole which is the sum of atomic masses i.e; C and 2 oxygen atoms

Molar mass of CO2 =12(C)+2*16(O) = 44 g/mole will contain 6.023 ※10^23 CO2 molecules ..

44g/mole = 6.023 ※10^23 CO2 molecules

=> 1g = (6.023/44) ※10^23 CO2 molecules

==> 8.80g = 8.80(6.023÷44)10^23 = 1.2046 ※10^23 molecules of CO2….

Thus there r 1.2046 ※10^23 molecules of CO2 in 8.80g

if u need to calculate no. of carbon atoms then multiply result by 1 and if u need no of oxygen atoms in 8.80g of co2 then multiply the result by 2 ….

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2 years ago

When a connector is marked with "al-cu," the connector is suitable for use with copper, copper-clad aluminum, and aluminum condu

IgorLugansk [536]

When connectors are marked with a combination of metals, it can be used as a connector of one of the metals or an alloy of the two metals. So in this case, since the marking is “Al – Cu” where Al is aluminium and Cu is copper, therefore the answer is:

<span>Yes, it is suitable for use with copper, copper-clad aluminum, and aluminum conductors.</span>

6 0

3 years ago

When 3243. grams of iron (III) chloride are reacted with 511.8 grams of hydrosulfuric acid, which is the limiting reactant?

AleksAgata [21]

Answer:

Hydrosulfuric acid will act as limiting reactant.

Explanation:

Given data:

Mass of iron(III) chloride = 3243.0 g

Mass of hydrosulfuric acid = 511.8 g

Limiting reactant = ?

Solution:

Chemical equation:

2FeCl₃ + 3H₂S → Fe₂S₃ + 6HCl

Number of moles of iron(III) chloride:

Number of moles = mass/molar mass

Number of moles = 3243.0 g/ 162.2 g/mol

Number of moles = 20 mol

Number of moles of hydrosulfuric acid:

Number of moles = mass/molar mass

Number of moles = 511.8 g/ 34.1 g/mol

Number of moles = 15 mol

Now we will compare the moles of both reactant with products

FeCl₃ : Fe₂S₃

2 : 1

20 : 1/2 ×20 = 10

FeCl₃ : HCl

2 : 6

20 : 6/2 ×20 = 60

H₂S : Fe₂S₃

3 : 1

15 : 1/3 ×15 = 5

H₂S : HCl

3 : 6

15 : 6/3 ×15 = 30

Hydrosulfuric acid producing less number of moles of product thus, it will act as limiting reactant.

5 0

3 years ago

List the following aqueous solutions in order of decreasing freezing point: 0.040 m glycerin (C3H8O3), 0.020 m potassium bromide

gogolik [260]

Answer:

Solutions from highest to lowest freezing point:

0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol

Explanation:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f$ = Depression in freezing point

i = van'T Hoff fcator

$K_f$ = Molal depression constant of solvent

m = molality of the solution

Higher the value of depression in freezing point at lower will be freezing temperature the solution.

1. 0.040 m glycerin

Molal depression constant of water = $K_f=1.86^oC/m$

i = 1 ( organic molecule)

m = 0.040 m

$\Delta T_{f,1}=1\times\times 1.86^oC/m\times 0.040 m$

$\Delta T_{f,1}=0.0744^oC$

2. 0.020 m potassium bromide

Molal depression constant of water = $K_f=1.86^oC/m$

i = 2 (ionic)

m = 0.020 m

$\Delta T_{f,2}=2\times\times 1.86^oC/m\times 0.020 m$

$\Delta T_{f,2}=0.0744^oC$

3. 0.030 m phenol

Molal depression constant of water = $K_f=1.86^oC/m$

i = 1 (organic)

m = 0.030 m

$\Delta T_{f,3}=1\times\times 1.86^oC/m\times 0.030 m$

$\Delta T_{f,3}=0.0558^oC$

$0.0744^oC=0.0744^oC > 0.0558^oC$

$\Delta T_{f,1}=\Delta T_{f,2}>\Delta T_{f,3}$

Solutions from highest to lowest freezing point:

0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol

5 0

3 years ago

Elements that readily lose electrons tend to have

babymother [125]

Answer:

C) low ionization energy and low electronegativity

Explanation:

8 0

3 years ago