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11Alexandr11 [23.1K]
3 years ago
10

This number tells you the number of each type of atom in a compound.

Chemistry
1 answer:
KonstantinChe [14]3 years ago
7 0

Answer:

avogadro's constant

Explanation:

this is the fixed number of the atoms in the molecule of an element

avogadro's law states that equal volumes of gases<em> </em><em>at</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>pressure</em><em> </em><em> </em><em>contain</em><em> </em><em>equal</em><em> </em><em>numbers</em><em> </em><em>of</em><em> </em><em>molecules</em><em> </em>

<em>that</em><em> </em><em>is</em><em> </em><em>all</em><em> </em><em>gases</em><em> </em><em>with</em><em> </em><em>same</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>pressure</em><em> </em><em>will</em><em> </em><em>always</em><em> </em><em>have</em><em> </em><em>equal</em><em> </em><em>numbers</em><em> </em><em>of</em><em> </em><em>molecules</em><em> </em>

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Use the Rydberg equation to calculate the wavelength (in nm) of the photon emitted when a hydrogen atom undergoes a transition f
natka813 [3]

Answer:

The answer is 434nm

Explanation:

The Rydberg equation is an empirical relationship expressed by Balmer and Rydberg which is stated as:

1/λ =      R_{H} (\frac{1}{n_{f} ^{2} }-\frac{1}{n_{i} ^{2} } ).............................. (1)

where λ is the wavelength, R_{H} is the Rydberg constant equal to 1.097 x 10^{7}m^{-1}, n is the transition level number, the subscript f and i are the final and initial levels respectively. Therefore for final transition n = 2 and for initial transition n = 5.

Making substitutions into equation (1), gives

1/λ =  1.097 x 10^{7} (\frac{1}{2^{2} }-\frac{1}{5^{2} })

   = 1.097 x 10^{7} (\frac{1}{4} - \frac{1}{25} )

     = 1.097 x 10^{7} (0.25 - 0.04)

   = 1.097 x 10^{7} x 0.21

   = 2303700

∴ λ = \frac{1}{2303700} = 4.34 x 10^{-7}m

Converting to nm, we have

  λ   = \frac{4.34 x 10^{-7} }{10^{-9} } = 434nm

Therefore, the wavelength of the emitted photon is 434nm

6 0
4 years ago
What reagent would distinguish between Ag+ and Fe3+?
Andreyy89
The reagent which is used is "NaI", that is Sodium iodide.
When the reagent sodium iodide is added to distinguish the Ag⁺ ions and Fe³⁺ ions, it will react and form precipitate with one, while have no precipitates formed with the other, so that is how we can distinguish both of them.
8 0
3 years ago
D ndjscns fenvffhv bvi uo
aleksandrvk [35]
Yes, very informational. thank you.
6 0
3 years ago
Read 2 more answers
An empty graduated cylinder weighs 55.26 g. When filled with 50.0 mL of an unknown liquid, it weighs
marysya [2.9K]

Answer:

.743 g/mL

Explanation:

Hopefully this helps :)

5 0
3 years ago
When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

  • K: 39.098;
  • N: 14.007;
  • O: 15.999.

Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

What's the enthalpy change of this process?

Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

4 0
3 years ago
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