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1 year ago

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Imagine you needed to identify if an object has undergone a physical change or a chemical change. What information would you nee

d to know, and how would you decide?

1 answer:

pishuonlain [190]1 year ago

7 0

How it looks. basically the thing that tells you how it change. for example if an ice cube was melted (heat), it only changed physically not chemically as the h20 molecules are still there. however lets say you burn woos— you cant get that would back. its ash now and it has changed chemically.

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Imagine you are a detective examining a crime scene. You are trying to

Sidana [21]

Answer:

C. pieces of hair found at the crime scene.

Explanation:

using the pieces of hair, you can get the DNA of the person. This will give u a better lead in solving the crime.

Hope it helps u ....

4 0

2 years ago

Read 2 more answers

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

1 year ago

How many moles are in 6.5e23 atoms of Ne?

Savatey [412]

<h3>Answer:</h3>

1.1 mol Ne

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 6.5 × 10²³ atoms Ne

[Solve] moles Ne

<u>Step 2: Identify Conversion</u>

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 6.5 \cdot 10^{23} \ atoms \ Ne(\frac{1 \ mol \ Ne}{6.022 \cdot 10^{23} \ atoms \ Ne})$
[DA] Divide [Cancel out units]: $\displaystyle 1.07938 \ mol \ Ne$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.07938 mol Ne ≈ 1.1 mol Ne

7 0

1 year ago

Which is an example of practical pursuit of alchemy

Oduvanchick [21]

A practical pursuit of alchemy was the development of metallurgy practices. As alchemists always tried to turn various metals into other things, metalworking techniques were developed even though their experiments were often unsuccessful.
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8 0

2 years ago

Read 2 more answers

The half-life of a positron is very short. It reacts with an electron, and the masses of both are converted to two gamma-ray pho

sp2606 [1]

Explanation:

(a) It is known that relation between energy and mass is as follows.

$E = 2 \times mc^{2}$

where, E = energy

m = mass

c = speed of light = $3 \times 10^{8}$ m/s

As it is given that mass is $9.109 \times 10^{-31}$ kg. So, putting the given values into the above formula as follows.

$E = 2 \times mc^{2}$

= $2 \times 9.109 \times 10^{-31} kg \times 3 \times 10^{8}m/s$

= $1.638 \times 10^{-13} J$

Therefore, we can conclude that the energy produced by the reaction between one electron and one positron is $1.638 \times 10^{-13} J$ .

(b) When gamma ray photons are produced then they will have the same frequency. Relation between energy and frequency is as follows.

E = $h \times \nu$ ..... (1)

where, h = plank's constant = $6.626 \times 10^{-34} J.s$

$\nu$ = frequency

Also, $E = 2 \times mc^{2}$ ........ (2)

Hence, equating equations (1) and (2) as follows.

$h \times \nu$ = $2 \times mc^{2}$

So,

$6.626 \times 10^{-34} Js \times \nu$ = $1.638 \times 10^{-13} J$

$\nu$ = $1.236 \times 10^{20} Hz$

Thus, we can conclude that the frequency is $1.236 \times 10^{20} Hz$ .

5 0

2 years ago