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oksian1 [2.3K]
3 years ago
5

The equation for the reaction between copper and nitric acid is

Chemistry
2 answers:
Stells [14]3 years ago
5 0
3Cu + 8HNO3 —— 3Cu(NO3)2 + 2NO + 4H2O
serious [3.7K]3 years ago
5 0

Answer:

v=1

w=4

x=1

y=2

z=2

Explanation:

Cu +4HNO₃ ⇒ Cu(NO₃)₂ + 2NO₂+ 2H₂O

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How many molecules in each sample?<br><br> 64.7 g N2<br> 83 g CCl4<br> 19 g C6H12O6
lilavasa [31]

Answer:

  • 1.39x10²⁴ molecules N₂
  • .25x10²³ molecules CCl₄
  • 6.38x10²² molecules C₆H₁₂O₆

Explanation:

First we <u>convert the given masses into moles</u>, using the <em>compounds' respective molar mass</em>:

  • 64.7 g N₂ ÷ 28 g/mol = 2.31 mol N₂
  • 83 g CCl₄ ÷ 153.82 g/mol = 0.540 mol CCl₄
  • 19 g C₆H₁₂O₆ ÷ 180 g/mol = 0.106 mol C₆H₁₂O₆

Then we multiply each amount by <em>Avogadro's number</em>, to <u>calculate the number of molecules</u>:

  • 2.31 mol N₂ * 6.023x10²³ molecules/mol = 1.39x10²⁴ molecules
  • 0.540 mol CCl₄ * 6.023x10²³ molecules/mol = 3.25x10²³ molecules
  • 0.106 mol C₆H₁₂O₆ * 6.023x10²³ molecules/mol = 6.38x10²² molecules
3 0
3 years ago
Which set of terms best defines what affects kinetic energy and potential energy, respectively?
Stolb23 [73]
I’d say velocity and height
3 0
4 years ago
If you have 8 g of a radioactive element there will be Of the radioactive element left after one half life
marta [7]

Answer:

4g of the radioactive element present

Explanation:

Radioactive elements are atoms that are unstable in normal conditions and are converted in another different atoms.

1 half life express the time that is need to decay the half of the initial amount of the radioactive element.

If initially we have 8g of the radioactive element, after 1 life we will have 8g/2 =

4g of the radioactive element present

3 0
3 years ago
Calculate the energy of light having a wavelength of 395 nm. <br> show work
Liono4ka [1.6K]

Answer: 5.032\times 10^{-19}\ J

Explanation:

Given

The wavelength of the light is \lambda =395\ nm

Energy associated with a photon(light) is given by E=\frac{hc}{\lambda}

where, h=\text{Planck constant}

c=\text{speed of light}

\Rightarrow E=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{395\times 10^{-9}}\\\\\Rightarrow E=5.032\times 10^{-19}\ J

Therefore, the energy associated with the light is 5.032\times 10^{-19}\ J

6 0
4 years ago
What property of a solution stays the same after its diluted
zepelin [54]
The amount of solute
5 0
3 years ago
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