<span>Determine the root-mean-square sped of CO2 molecules that have an average Kinetic Energy of 4.21x10^-21 J per molecule. Write your answer to 3 sig figs.
</span><span>
E = 1/2 m v^2
If you substitute into this formula, you will get out the root-mean-square speed.
If energy is Joules, the mass should be in kg, and the speed will be in m/s.
1 mol of CO2 is 44.0 g, or 4.40 x 10^1 g or 4.40 x 10^-2 kg.
If you divide this by Avagadro's constant, you will get the average mass of a CO2 molecule.
4.40 x 10^-2 kg / 6.02 x 10^23 = 7.31 x 10^-26 kg
So, if E = 1/2 mv^2
</span>v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 115184.68
Take the square root of that, and you get the answer 339 m/s.
Answer: 1.027 x 10^6 g= 1027kg
In this question, you are given the volume of the blimp (2.027×10^5 ft^3) and the density of the gas(0.179g/L). To answer this question, you need to convert the volume unit into liter. The calculation would be: 2.027×10^5 ft^3 x 28.3168L/ft3= 57.398 x 10^5L= 5.74x10^6L
Then to find the mass, multiply the volume with the density. The calculation would be: 5.74x10^6L x 0.179g/L= 1.027 x 10^6 g= 1027kg
Answer:
333.7g of antifreeze
Explanation:
Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:
ΔT = Kf × m × i
Where:
ΔT is change in temperature (0°C - -20°C = 20°C)
Kf is freezing point depression constant (1.86°C / m)
m is molality of solution (moles solute / 0.5 kg solvent -500g water-)
i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)
Replacing:
20°C = 1.86°C / m × moles solute / 0.5 kg solvent × 1
5.376 = moles solute
As molar mass of ethylene glycol is 62.07g/mol:
5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.
1. sundering
2. talc
3 luster
Answer: The correct name for the compound 
is, Dicarbon triiodide.
Explanation:
is a covalent compound because in this compound the sharing of electrons takes place between carbon and iodine.. Both the elements are non-metals. Hence, it will form covalent bond.
The naming of covalent compound is given by:
The less electronegative element is written first.
The more electronegative element is written second. Then a suffix is added with it. The suffix added is '-ide'.
If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms and so on.
Hence, the correct name for the compound is, Dicarbon triiodide..
