Answer:
1.33 atm
Explanation:
use general gas equation P1 V1/ T1 = P2 V2/ T2
rearrange and make P2 the subject then solve,it should give you 1.33 atm
Covalent Bond Name :N6Cl10
Explanation:
- Covalent compounds are named by using numerical prefixes to identify the number of atoms in the molecule. For example Carbon Dioxide CO2 and Carbon Monoxide CO . ... Drop the double vowel for the prefix and the element of the second element in the compound.
- The three types as mentioned in the other answers are polar covalent, nonpolar covalent, and coordinate covalent. The first, polar covalent, is formed between two nonmetals that have a difference in electronegativity. They share their electron density unevenly.
- Examples of Covalent Bond: Water. An example is water. Water consists of a covalent bond containing hydrogen and oxygen bonding together to make H2O. ...Diamonds. A diamond is an example of Giant Covalent bond of carbon. A diamond has a giant molecular structure. ...Vulcanized rubber. Another example is vulcanized rubber.
- Covalent bonds are especially important since most carbon molecules interact primarily through covalent bonding. Covalent bonding allows molecules to share electrons with other molecules, creating long chains of compounds and allowing more complexity in life.
Answer:
The molar mass of copper (II) nitrate is 187.5 g/mol.
Explanation:
The molar mass is the mass of all the atoms in a molecule in grams per mole. To calculate the molar mass of a molecule, we first obtain the atomic weights from the individual elements in a periodic table. We then count the number of atoms and multiply it by the individual atomic masses.
Answer:
2,909 M
Explanation:
molair mass is of.ethylene is 26,04 g/mol
first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.
3000 g x 1.1 = 3300 mL = 3,3 L
Next you need to calculate the amount of moles:
250 g / 26,04 g/mol = 9,60 mol
Now you can calculate the molarity:
9,6/3.3 = 2,909 M
I don't know the answer for the second question. I'm sorry.
