Explain whether co2 (o = c = o) is an ionic or a covalent compound. why does this arrangement satisfy all atoms involved?
1 answer:
Carbon dioxide (CO2) is a covalent compound because it is composed of carbon and oxygen. Herein both the elements i.e. C and O are non-metals.
These elements share the electrons. When electrons are shared, each atom will contain 8 electrons in the outer shell. Due to this, all the atoms of the CO2 will become stable and they will have a completed outer shell.
The lewis structure of CO2 is shown below:
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Answer:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Explanation:
By ideal gas equation:
Number of moles (n)
can be written as:
where, m = given mass
M = molar mass
where,
which is known as density of the gas
The relation becomes:
.....(1)
We are given:
M = molar mass of chloroform= 119.5 g/mol
R = Gas constant =
T = temperature of the gas =
P = pressure of the gas = 1.00 atm
Putting values in equation 1, we get:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Answer:
Explanation:
To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.
1. Convert Atoms to Moles
To convert atoms to grams, Avogadro's number must be used.
This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.
Multiply the given number of atoms by Avogadro's number.
Flip the fraction so the atoms of iodine will cancel out.
Multiply so the problem condenses into 1 fraction.
2. Convert Moles to Grams
Now we must use the molar mass of iodine, which is found on the Periodic Table.
- Iodine Molar Mass: 126.9045 g/mol
Use this mass as a fraction.
Multiply this fraction by the number of moles found above.
Multiply. The moles of iodine will cancel.
The 1 as a denominator is insignificant.
3. Round
The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.
The 8 in the hundredth place tells us to round the 0 up to a 1.
There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.