First, change the 39.27grams of water to moles using the molar mass.
Molar mass H2O= 18.02 grams/mole
39.27 grams x (1 mole H2O/ 18.02 grams)= 2.18 moles
The mass of sodium acetate can be calculated by subtracting the mass of the compound minus the mass of water.
84 g - 39.27 grams= 44.73 grams
Now change does to moles using the molar mass
Molar mass of sodium acetate= 82 g/ mol
44.73 g x (1 mole/ 82 grams) = 0.545 mole
Now divide the both value by the smallest .
H2O- 2.18/ 0.545= 4
sodium acetate= 0.545/ 0.545= 1
The formula is NaC2H3O2 • 4 H2O
B. The unknown solution had the lower concentration.
Explanation:
Osmosis is a phenomenon in which the molecules of the solvent has a tendency to move through a membrane which is semipermeable from lower concentrated side to the higher concentration side, so that the concentrations on both sides of the membrane must be equal.
So the unknown solution may have lesser concentration than the isotonic solution so that molecules of that solution move from less concentrated side to the more concentrated side, so its level drops.
Answer:
hydrogen ions
Explanation:
because acid is the specie that have ability to donate proton or forming bond with electron pair
Answer:
4.34x10⁻¹⁹ J
Explanation:
The total energy emitted by irradiation is given by
E = hf, where E is the energy, <em>h</em> is the plack constant (6.626x10⁻³⁴ J.s), and <em>f</em> is the frequency. The frequency is also the velocity of the light (c = 2.99x10⁸ m/s) divided by the length of the irradiation (254x10⁻⁹ m). So:
E = (6.626x10⁻³⁴)x(2.99x10⁸)/ (254x10⁻⁹)
E = 7.80x10⁻¹⁹ J
The energy to remove 1 electron is the energy necessary to remove 1 mol divided by the Avogadros number ( 1 mol = 6.02x10²³ electrons):
208400/6.022x10^23 = 3.46x10⁻¹⁹ J
The total energy is the energy necessary to remove one electrons plus the kinectic energy (Ek) of the electrons:
7.80x10⁻¹⁹ = 3.46x10⁻¹⁹ + Ek
Ek = 7.80x10⁻¹⁹ - 3.46x10⁻¹⁹
Ek = 4.34x10⁻¹⁹ J
1 ounces ----------- 29.5735 mL
5 ounces ----------- ??
5 x 29.5735 / 1 => 147.868 mL
hope this helps!
