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velikii [3]
3 years ago
9

How many particles are in 12.47 grams of NaCl?

Chemistry
1 answer:
jeka943 years ago
5 0
If we are talking about moles then the answer to that is 0.22
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Can anyone do this question.i really need a answer​
PilotLPTM [1.2K]

Answer:

17.4

Explanation:

kasi 15.2

16.2

=17.4 thanks me later pa brainliest din

7 0
3 years ago
Homogeneous series of n-propyl bromide?
goldenfox [79]

The homologous series ( not homogeneous series ) of n-propyl bromide is C3H7Br. It is also known as 1-Bromopropane

n-propyl bromide or 1-bromopropane, is a solvent that is used in cleaning metals, vapor degreasing and also for dry cleaning.

<h3>What is Homologous series?</h3>

Homologous series is a family of organic compound which follows a regular structural pattern and in which successive members differs from one another by a molecular formula of CH2

Below are some of their characteristics:

  • The general formula of all compounds in the series is the same.
  • They have the same functional group.
  • Their physical properties such as melting point, boiling point, density, generally show a gradual change with increase of molecular formula in the series.

Learn more about homogeneous series:

brainly.com/question/14008526

3 0
2 years ago
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Complete the equation for respiration: Oxygen + ________ ---&gt; carbon dioxide + energy + water *
Tom [10]

Answer:Sunlight

Explanation:

5 0
3 years ago
Help pleaseeeeeeeeeeeeee!!?
melisa1 [442]

Answer:

There are 146 neutrons in Uranium-238

There are 90 electrons in Thorium-234

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Electric power is measured in watts (1 W = 1 J/s). About 95% of the power output of an incandescent bulb is converted to heat an
nlexa [21]

Answer:

9.430 * 10¹⁷ protons per second whill shine on the book from a 62 W bulb

Explanation:

To answer this question, first let's calculate the energy of a single photon with a wavelength (λ) of 504 nm:

E = hc/λ

Where h is Planck's constant (6.626*10⁻³⁴ J·s) and c is the speed of light (3*10⁸ m/s).

E = 6.626*10⁻³⁴ J·s * 3*10⁸ m/s ÷ (504*10⁻⁹m) = 3.944 * 10⁻¹⁹ J.

So now we can make the equivalency for this problem, that

<u>1 proton =  3.944 * 10⁻¹⁹ J</u>

Now we convert watts from J/s to proton/s:

1 \frac{J}{s}*\frac{1proton}{3.944*10^{-19}J}=2.535*10^{18} \frac{proton}{s} = 1 W

Solving the problem, a 62 W bulb converts 5% of its output into light, so:

  • 62 * 5/100 = 3.1 W

3.1 watts are equal to [ 2.535*10¹⁸ proton/s * 3.1 ] = 7.858 * 10¹⁸ proton/s

Of those protons per second, 12% will shine on the chemistry textbook, thus:

7.858 * 10¹⁸ proton/s * 12/100 = 9.430 * 10¹⁷ protons/s

8 0
2 years ago
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