Answer:
Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of
is larger than that of
(by a factor of about
.) Therefore, the mass of the
sample is significantly larger than that of the
sample.
Explanation:
The
and the
sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles (
and
molecules, respectively.) That is:
.
Note that the mass of a gas
is different from the number of gas particles
in it. In particular, if all particles in this gas have a molar mass of
, then:
.
In other words,
.
.
The ratio between the mass of the
and that of the
sample would be:
.
Since
by Avogadro's Law:
.
Look up relative atomic mass data on a modern periodic table:
Therefore:
.
.
Verify whether
:
- Left-hand side:
. - Right-hand side:
.
Note that the mass of the
sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.
Answer:
Sodium Hydroxide + Sulfuric Acid = Sodium Sulfate + Water
2NaOH + H2SO4 → Na2SO4 + 2H2O
Explanation:
To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. The balanced equation will appear above.
Use uppercase for the first character in the element and lowercase for the second character. Examples: Fe, Au, Co, Br, C, O, N, F.
Ionic charges are not yet supported and will be ignored.
Replace immutable groups in compounds to avoid ambiguity. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will.
Compound states [like (s) (aq) or (g)] are not required.
You can use parenthesis () or brackets [].
Answer:
-0.1767°C (Option A)
Explanation:
Let's apply the colligative property of freezing point depression.
ΔT = Kf . m. i
i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1
m = molality (mol of solute / 1kg of solvent)
We have this data → 0.095 m
Kf is the freezing-point-depression constantm 1.86 °C/m, for water
ΔT = T° frezzing pure solvent - T° freezing solution
(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1
T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C
Answer: 0.08695652
Explanation:
You would do the answer you got subtracting from the expected answer over your expected answer
Answer:
water, when the metastable state is reached, is cooled below the zero temperature. It freezes abruptly. this is called metastable. They are not at equilibrium per se; as at negative temperatures the only equilibrium state of water is ice.
Explanation: