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3 years ago

34. Explain how dimensional analysis is used to solve problems.

Chemistry

1 answer:

alisha [4.7K]3 years ago

7 0

Answer: By understanding conversion factors and how they are related to each other

Explanation:

Dimensional Analysis is a step by step approach to solving problems in Physics, Chemistry , and Mathematics. It involves having a clear knowledge and understanding to be able to convert a given unit to another in the same dimension using conversion factors and knowing how they are related to each other.

For instance, In Chemistry, we want to Convert 120mL to L.(note that ml stands for millilitres and ;L stands for litres)

Or first approach will be to write out the conversion factor related to our problem which is

1000ml =1L

such that 120ml = (we cross multiply))

giving us 120ml x 1L/1000ml =0.12L

This same process is applied to convert any type of dimensional analysis problems be it physics or mathematics.

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Convert the volume 0.455 m3 to the unit mL

Stells [14]

The answer is 455000

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4 years ago

Help on science please!! Will bark brainliest!

jok3333 [9.3K]

Aphelion in miles =94290301

Perihelion in miles = 91200281

Explanation:

To convert km in mile we use the multiply the distance in km with 0.62, because there are 0.62 miles in 1 km.

Aphelion in miles = aphelion in km × 0.62

Aphelion in miles = 152081132 × 0.62

Aphelion in miles =94290301

Perihelion in miles = perihelion in km × 0.62

Perihelion in miles = 147097229 × 0.62

Perihelion in miles = 91200281

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3 years ago

How many moles of N₂ are essential for generating 0.08 moles of Li₃N in the given reaction?

Murrr4er [49]

For every 2 moles of Li3N, you need 1 mole of N2
so for 0.08 mol Li3N, you need 0.04 moles N2

3 0

3 years ago

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When 108 g of water at a temperature of 23.9 °c is mixed with 66.9 g of water at an unknown temperature, the final temperature o

olganol [36]

Here,

Heat gain by the first sample of water + Heat lost by the second sample of water is equal to zero (0).

Now, Mass of water sample one = 108 g (given)

Mass of water sample two = 66.9 g (given)

Temperature for water sample one = $23.9^{0}C$

Let, temperature for water sample two =x

And, final temperature = $47.2^{0}C$

Now,

$mass of water sample one\times specific heat of water\times (T_{f} - T_{i}) + mass water sample two\times specific heat of water\times (T_{f} - T_{i}) = 0$

where, $T_{f}$ = final temperature

$T_{i}$ = initial temperature

Substitute all the given values in above formula:

$(108 g\times 4.184 J/g . K\times ( 47.2^{0}C- 23.9^{0}C) )+ (66.9 g \times 4.184 J/g . K\times (47.2^{0}C -x))= 0$

$(451.872 J/K \times (23.3^{0}C)) + (279.9096 J/K\times (47.2^{0}C -x)) = 0$

$(10528.6176 J/K(^{0}C)+ (13211.73312 J/K(^{0}C) -279.9096 J/K\times x)= 0$

$(23740.35072 J/K(^{0}C) -279.9096 J/K\times x)= 0$

$-279.9096 J/K\times x= -23740.35072 J/K(^{0}C)$

$x =\frac{23740.35072 J/K(^{0}C)}{279.9096 J/K}$

[tex x =84.81^{0}C [/tex]

5 0

4 years ago

Give the theoretical yield, in moles, of CO2 from the reaction of 4.00 moles of C8H18 with 4.00 moles of O2. 2 C8H18(l) + 25 O2(

scoray [572]

2C₈H₁₈ + 25O₂ = 16CO₂ + 18H₂O

n(C₈H₁₈)=4 mol
n(O₂)=4 mol

C₈H₁₈:O₂=2:25 ⇒ 4:50 oxygen is the limiting reagent

n(CO₂)=16n(O₂)/25

n(CO₂)=16*4/25=2.56 mol

6 0

3 years ago

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