Bonds of two atoms of equal electronegativity are nonpolar covalent bonds.
Your second sentence is identical to the first sentence; I'll bet the second sentence is "Bonds between two atoms that are unequally electronegative are polar covalent bonds."
<span>At standard temperature and pressure 22.4 l of an ideal gas would contain 1 mole. in order to find the change in moles we must look at the ideal gas law PV=nRT where P=Pressure V=volume n=Moles R= Gas constant T= Temperature. To simplify this equation we will be using the gas constant at .08206 L-atm/mol-K. We must first convert 100c to k which is 373.15. Then we can plug the values into our equation which gives us (2atm)(14.5 l)=(n)(.08206 L-atm/mol-K)(373.15). After some basic algebra we get the moles to equal roughly .95 which is .05 moles less than our original system.</span>
Answer: The concentration of excess ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
in solution is 0.017 M.
Explanation:
1. 
moles of 
1 mole of 
give = 1 mole of 
Thus 0.019 moles of give = 0.019 mole of
2. moles of 
According to stoichiometry:
1 mole of 
gives = 2 moles of 
Thus 0.012 moles of give = 
moles of
As 1 mole of neutralize 1 mole of
0.019 mole of will neutralize 0.019 mole of
Thus (0.024-0.019)= 0.005 moles of will be left.
![[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B%5Ctext%20%7Bmoles%20left%7D%7D%7B%5Ctext%20%7BTotal%20volume%20in%20L%7D%7D%3D%5Cfrac%7B0.005%7D%7B0.3L%7D%3D0.017M)
Thus molarity of in solution is 0.017 M.
Answer:
1.505×10^23 atoms of lead
Explanation:
Volume of lead in the lungs = total volume of lungs = 5.60L
1 mole = 22.4L
5.6L of lead = 5.6/22.4 = 0.25 mole
From Avogadro's law
1 mole of lead contains 6.02×10^23 atoms of lead
0.25 mole of lead = 0.25×6.02×10^23 = 1.505×10^23 atoms of lead
Answer:
Calculate the mass percent of a potassium nitrate solution when 15.0 g KNO3 is dissolved in 250 g
of water.
2. Calculate the mass percent of a sodium nitrate solution when 150.0 g NaNO3 is dissolved in 500 mL
of water. Hint: 1 mL water = 1 g water
3. Calculate the weight of table salt needed to make 670 grams of a 4.00% solution.
4. How many grams of solute are in 2,200 grams of a 7.00% solution?
5. How many grams of sodium chloride are needed to prepare 6,000 grams of a 20% solution?
Mass Percent = Grams of Solute
Grams of Solution X 100%
100%
Grams of Solute = Grams of Solution X Mass Percent
= 26.8 grams NaCl
= 670 grams X 4.00%
100%
100%
Grams of Solute = Grams of Solution X Mass Percent
= 154 grams solute
= 2,200 grams X 7.00%
100%
100%
Grams of Solute = Grams of Solution X Mass Percent
= 1,200 grams NaCl
= 6,000 grams X 20.0%
100%
Explanation:
