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3 years ago

BRAINLIEST WILL BE GIVEN, PLZ HELP + 15 PTS

Chemistry

2 answers:

podryga [215]3 years ago

7 0

Answer:

The correct answer is the last option

Explanation:

zepelin [54]3 years ago

4 0

A I just took the test

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How much heat energy is required to raise the temperature of 0.358 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co

NNADVOKAT [17]

In order to calculate how much heat is needed to raise the temperature you need to use the formula q =mass x specific heat x (final temperature- initial temperature) where q represents heat being absorbed or released. Before you begin you would convert kg to g because the specific heat is measure in g. So you would set up the equation as q = 358 g x .092 x (60-23 degrees Celsius) which would give you 1218.6

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4 years ago

Read 2 more answers

C5H8+ _02 = 8h2o+5co2

Dimas [21]

To balance this equation you would do whatever you did I TO one side to the other... so if there is 18 oxygen on the right side then there must be 18 on the left. So 18/2 equals 9... the. You equation will be balanced

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3 years ago

If you observe bubbles forming in a mixture, what is happening?

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4 years ago

Given the wavelength of the corresponding emission line, calculate the equivalent radiated energy from n = 4 to n = 2 in both jo

lions [1.4K]

Explanation:

It is given that,

Initial orbit of electrons, $n_i=4$

Final orbit of electrons, $n_f=2$

We need to find energy, wavelength and frequency of the wave.

When atom make transition from one orbit to another, the energy of wave is given by :

$E=-13.6(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

Putting all the values we get :

$E=-13.6(\dfrac{1}{(4)^2}-\dfrac{1}{(2)^2})\\\\E=2.55\ eV$

We know that : $1\ eV=1.6\times 10^{-19}\ J$

So,

$E=2.55\times 1.6\times 10^{-19}\\\\E=4.08\times 10^{-19}\ J$

Energy of wave in terms of frequency is given by :

$E=hf$

$f=\dfrac{E}{h}\\\\f=\dfrac{4.08\times 10^{-19}}{6.63\times 10^{-34}}\\\\f=6.14\times 10^{14}\ Hz$

Also, $c=f\lambda$

$\lambda$ is wavelength

So,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{6.14\times 10^{14}}\\\\\lambda=4.88\times 10^{-7}\ m\\\\\lambda=488\ nm$

Hence, this is the required solution.

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3 years ago

The compound adrenaline contains 56.7 % c, 6.56 % h, 28.37% o and 8.28 % n by mass. what is the empirical formula for adrenaline

-Dominant- [34]

Answer:- An empirical formula of adrenaline is $C_8H_1_1O_3N$ .

Solution:- Convert given percentages to moles and then calculate the mol ratio to get the simplest whole number ratio of moles of atoms which is the empirical formula of the compound.

$C=\frac{56.7}{12}$ = 4.725 mol