Explanation:
<h3 /><h2>
<em><u>H2 </u></em><em><u>+</u></em><em><u> </u></em><em><u>O2 </u></em><em><u>=</u></em><em><u> </u></em><em><u>H2O</u></em></h2>
<h2>
<em><u>Hydrogen</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>Oxygen</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>Water</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em></h2>
<em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em>
The false statement from the above is that: Temporary charge imbalances in the molecules lead to London dispersion forces.
<h3>What are the factors that affect London dispersion forces?</h3>
Generally, the factors which affects the London dispersion forces a dispersion force are as follows:
- Shape of the molecules
- Distance between molecules
- Polarizability of the molecules
However, London dispersion forces simply refers to a sort of temporary attractive force formed when electrons in two adjacent atoms occupy positions that make the atoms form dipoles.
So therefore, temporary charge imbalances in the molecules lead to London dispersion forces is a false statement
Learn more about London dispersion forces:
brainly.com/question/1454795
Density is proportional to molar mass, assuming pressure and
temperature remain constant. Therefore, since CO has a molar mass of 28
and CO2 has a molar mass of 44:
The relative density of CO vs air is 28/29 = 0.9655.
The relative density of CO2 vs air is 44/29 = 1.517.
Explanation:
Molarity = mol/L or g/L
Data:
Mass>80.0g
Mr of CaCl2>111.1g/mol
V>500mL
Convert mL to Litres;. mole=g/Mr
1L=1000mL. =80g/111.1g/mol
x=500mL. =0.720072007mol
x=0.5L
Molarity= mole/volume
=0.720072007mol/0.5L
=1.440144014mol/L
=1.4401mol/L
Hope this helps. Depending on the question you can also find it in g/L but mol/L is safer.
Answer:
<em>To calculate the average atomic mass, multiply the fraction by the mass number for each isotope, then add them together.</em>
<em>hope this helps</em><em> </em><em><</em><em>3</em>
