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2 years ago

create a flowchart that outlines when to use dimensional analysis and when to use scientific notation.

Chemistry

1 answer:

murzikaleks [220]2 years ago

7 0

Answer:

see attached

Explanation:

Dimensional analysis is useful whenever dimensions are involved. Unless it is quite clear that all of the problem dimensions are consistent (for example, all speeds in miles per hour, or all angles in degrees), dimensional analysis can be useful for keeping the math straight.

Only units of the same dimensions can be added or subtracted. When numbers are multiplied or divided or raised to a power, dimensional analysis can help ensure that the appropriate operations are being used on appropriate numbers. It can also help ensure that dimensions are being combined properly to give appropriate derived dimensions.

Scientific notation is a way of writing very large or very small numbers compactly. It can also help with "order of magnitude" estimates. If an answer using SI prefixes is appropriate, or if a number can be conveniently expressed in standard form, then scientific notation is usually not required.

On the other hand, SI prefixes may not be appropriate in some cases, or a problem may specify that scientific notation be used for expressing results. In those instances, scientific notation should be used.

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3) After a rather lengthy organic chemistry synthesis procedure, a student ran the product of the reaction on a TLC plate and obtained the result below. What might he/she have done wrong, if anything?

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The information on these pages is available for academic use without restriction.

3 0

1 year ago

The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera

marshall27 [118]

Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2 cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius

8 0

1 year ago

Jeff wants to find the volume of a gas in a cylinder. He knows that at 28°C and a pressure of 1 atm, the volume of the gas was 6

emmasim [6.3K]

Answer:

by using ideal gas law

Explanation:

ideal gas law:

PV=nRT

where:

P is pressure measured in Pascal (pa)

V is volume measured in letters (L)

n is number of moles

R is ideal gas constant

T is temperature measured in Kelvin (K)

by applying the given:

P(initial) V(initial)=nRT(initial)

P(final) V(final)=nRT(final)

nR is constant in both equations since same gas

then,

P(initial) V(initial) / T(initial) = P(final) V(final) / T(final)

then by crossing multiply both equations

V (final)= { (P(initial) V(initial) / T(initial)) T(final) } /P (final)

P(initial)=P(final)= 1 atm = 101325 pa

V(initial)= 6 L

T(initial) = 28°c = 28+273 kelvin

T(final) = 39°c = 39+273 kelvin

by substitution

V(final) = 6.21926 L

6 0

1 year ago

Read 2 more answers

Air in a 0.3 m3 cylinder is initially at a pressure of 10 bar and a temperature of 330K. The cylinder is to be emptied by openin

sergiy2304 [10]

Answer:

(a) Temperature = 330 K, and mass = 0.321 kg

(b) T₂ = 171.56 K, mass = 0.32223 kg

Explanation:

For a constant temperature process we have

p₁v₁ = p₂v₂

Where p₁ = initial pressure = 10 bar = 1000000 Pa

p₂ = final pressure = 1 atm = 101325 Pa

v₁ initial volume = 0.3 m³

v₂ = final volume = unknown

From the relation we have v₂ = 2.96 m³

Therefore at constant temperature 2.93 m³ - 0.3 m³ or 2.66 m³ will be expelled from the container

Temperature = 330 K, and mass =

Also from the relation p1v1 = mRT1

We have, (1000000×0.3)/(8314×330) = 109..337 mole

For air mass

Mass = 3.171 kg

After opening we have

p2v2/(RT1) = n2 = 11.07 mol or 0.321 kg

(b) This is said to be adiabatic condition hence

Here

But cp = 29 (J/mol K).

and p₁v₁ = RT₁ therefore R = 1000000*0.3/330 = 909.1 J/mol·K

And For perfect gas γ = 1.4

Hence T₂ = 171.56 K

γ =cp/cv therefore cv=cp/γ = 29/1.4 = 20.714 (J/mol K). and R =cp-cv = 8.29 J/mol·K

Therefore p1v1/(RT1) = 109.66 moles and we have

p2v2/(R×T2) = 11.11 mole left

For air that is 0.32223 kg

5 0

2 years ago

A 251 g strip of glass wool is used to insulate a reaction flask. During the reaction the temperature of the glass wool increase

Ivahew [28]

Answer:

8.9 KJ

Explanation:

Given data:

Mass of strip = 251 g

Initial temperature = 22.8 °C

Final temperature = 75.9 °C

Specific heat capacity of granite = 0.67 j/g.°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 75.9 °C - 22.8 °C

ΔT = 53.1 °C

Q = 251 g × 0.67 j/g.°C × 53.1 °C

Q = 8929.8 J

Jolue to KJ.

8929.8J ×1 KJ / 1000 J

8.9 KJ

8 0

1 year ago