Answer:
The molarity of the solution: 0,27M
Explanation:
First , we calculate the weight of 1 mol of NaCl:
Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol
58,5 g---1 mol NaCl
64 g--------x= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl
A solution molar--> moles of solute in 1 L of solution:
4 L-----1,09 mol NaCl
1L----x0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M
Answer:
1.2 x 10⁵ moles Ag (2 sig. figs.)
Explanation:
1 mole any substance (elements or compounds) => 6.023 x 10²³ particles of specified substance
∴ 6.9 x 10²⁸ atoms Ag = 6.9 x 10²⁸ Ag atoms / 6.023 x 10²³ Ag atoms/mole Ag
= 1.145608501 x 10⁵ moles Ag (calculator answer)
= 1.2 x 10⁵ moles Ag (2 sig. figs.)
Answer:
The oxidation number of C (carbon) is +4
Explanation:
Answer:
Solute - The solute is the substance that is being dissolved by another substance. In the example above, the salt is the solute. Solvent - The solvent is the substance that dissolves the other substance.
Explanation:
Answer:
Explanation:
<u>1) Data:</u>
a) M = ?
b) mass of solue = 17 g
c) solute: NH₃
d) V = 0.5o liter
<u>2) Formulae:</u>
a) number of moles, n = mass in grams / molar mass
b) M = n / V (in liters)
<u>3) Solution</u>
a) Molar mass of NH₃ = 17.03 g/mol
b) n = mass in grams / molar mass = 17 g / 17.03 g/mol = 0.998 mol NH₃
c) M = n / V (in liters) = 0.998 mol / 0.50 liter = 1.996 M
d) Round to the appropiate number of significant figures, 2: 2.0 M.
Answer: 2.0 M
