Answer:the number of moles represented by 3.0 x 10^24 atoms of Ag is 0.500mol 0.500 m o l .
Explanation:
Answer:
★ Molecular geometry is described by VSEPR theory, which basically states that electron pairs around a central atom will repel each other, and get as far apart as possible, in three dimensions.
Explanation:
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Answer:
510 g NO₂
General Formulas and Concepts:
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
- Reading the Periodic Table
- Writing Compounds
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)
<u>Step 2: Define conversions</u>
Avogadro's Number
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol
<u>Step 3: Use Dimensional Analysis</u>
<u />
= 511.901 g NO₂
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules.</em>
511.901 g NO₂ ≈ 510 g NO₂
Answer:- Frequency is
.
Solution:- frequency and wavelength are inversely proportional to each other and the equation used is:
![\nu =\frac{c}{\lambda }](https://tex.z-dn.net/?f=%5Cnu%20%3D%5Cfrac%7Bc%7D%7B%5Clambda%20%7D)
where,
is frequency, c is speed of light and
is the wavelength.
Speed of light is
.
We need to convert the wavelength from nm to m.
(
)
![550nm(\frac{10^-^9m}{1nm})](https://tex.z-dn.net/?f=550nm%28%5Cfrac%7B10%5E-%5E9m%7D%7B1nm%7D%29)
= ![5.5*10^-^7m](https://tex.z-dn.net/?f=5.5%2A10%5E-%5E7m)
Now, let's plug in the values in the equation to calculate the frequency:
![\nu =\frac{3.0*10^8m.s^-^1}{5.5*10^-^7m}](https://tex.z-dn.net/?f=%5Cnu%20%3D%5Cfrac%7B3.0%2A10%5E8m.s%5E-%5E1%7D%7B5.5%2A10%5E-%5E7m%7D)
=
or ![5.4*10^1^4Hz](https://tex.z-dn.net/?f=5.4%2A10%5E1%5E4Hz)
since, ![1s^-^1=1Hz](https://tex.z-dn.net/?f=1s%5E-%5E1%3D1Hz)
So, the frequency of the green light photon is
.
Answer:
![m_{AgNO_3}=577.6mg](https://tex.z-dn.net/?f=m_%7BAgNO_3%7D%3D577.6mg)
Explanation:
Hello there!
In this case, according to the given chemical reaction, it is first necessary to compute the moles of reacting LiOH given its molar mass:
![n_{LiOH}=81.5mg*\frac{1g}{1000mg}*\frac{1mol}{23.95g} =0.0034molLiOH](https://tex.z-dn.net/?f=n_%7BLiOH%7D%3D81.5mg%2A%5Cfrac%7B1g%7D%7B1000mg%7D%2A%5Cfrac%7B1mol%7D%7B23.95g%7D%20%20%3D0.0034molLiOH)
Thus, since there is a 1:1 mole ratio between lithium hydroxide and silver nitrate (169.87 g/mol) the resulting milligrams turn out to be:
![m_{ AgNO_3}=0.0034molLiOH*\frac{1molAgNO_3}{1molLiOH} *\frac{169.87gAgNO_3}{1molAgNO_3} *\frac{1000gAgNO_3}{1gAgNO_3} \\\\m_{AgNO_3}=577.6mg](https://tex.z-dn.net/?f=m_%7B%20AgNO_3%7D%3D0.0034molLiOH%2A%5Cfrac%7B1molAgNO_3%7D%7B1molLiOH%7D%20%2A%5Cfrac%7B169.87gAgNO_3%7D%7B1molAgNO_3%7D%20%2A%5Cfrac%7B1000gAgNO_3%7D%7B1gAgNO_3%7D%20%5C%5C%5C%5Cm_%7BAgNO_3%7D%3D577.6mg)
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