Answer:
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Answer:
Molarity = 0.002 M
Explanation:
Given data:
Mass of calcium chloride = 0.321 g
Volume of water = 1.45 L
Molarity of solution = ?
Solution:
Molarity = number of moles / volume in litter.
We will calculate the number of moles of calcium chloride first.
Number of moles = mass/molar mass
Number of moles = 0.321 g/ 110.98 g/mol
Number of moles = 0.003 mol
Molarity:
Molarity = 0.003 mol / 1.45 L
Molarity = 0.002 M
Answer:
B) The molecular orbital formed is lower in energy than a hydrogen 1s atomic orbital.
Explanation:
When two atoms of hydrogen come close to each other , there is formation of molecular orbital . Due to overlap of 1 s orbital of one and 1 s orbital of another atom , two molecular orbitals are formed . One of these molecular orbital has energy less than 1 s atomic orbital . It is called 1 s sigma bonding molecular orbital . The other molecular orbital has energy more than 1 s atomic orbital . It is called antibonding molecular orbital . Two electrons occupy bonding sigma molecular orbital .
So , the statement that "the molecular orbital formed is lower in energy than a hydrogen 1s atomic orbital " is wrong .
Molar mass is the amount grams that one mole weighs.
Explanation: You need to find the molar mass of NaCl which is the same as the amu on the periodic table in grams. So it is 22.99(Na) + 35.45(Cl) = 58.44
You also know that for every mole of NaCl you have 1 mole of Na because every molecule of NaCl has 1 atom of Na.
Finally, using the periodic table, again, you see that the molar mass of Na is 22.99.
Then using stoichiometry, you can find the grams of sodium.
100(g NaCl) * 1 mol (NaCl)/58.44 g (NaCl) * 1 mol (Na)/ 1 mol (NaCl) * 22.99 (g of Na)/ 1 mol (Na)
which equals 39.339435 g of Na.
If you need to maintain significant figures the answer will be 40.
hope this helps
This is a combustion reaction. Any reaction in which a hydrocarbon decomposes into carbon dioxide and water is classified as combustion. The energy used for the decomposition usually comes from intense heat, which is why combustion of a hydrocarbon almost always is associated with a source of fire.
