Density is calculated as mass divided by volume. If we are given an ice cube of side length 8.00 cm, then the volume of the cube is equivalent to (8.00 cm)^3 = 512 cm^3. Since we have a given mass of 476 g, we can divide:
476 g / 512 cm^3 = 0.930 g/cm^3
So the density of the sample of ice is 0.930 g/cm^3.
Methane.
Water - H2O
Methane - CH4
Methane has 2 more hydrogens than water.
Answer : The molar mass of the solute would be low.
Explanation :
Formula used for depression in freezing point is:
where,
= change in freezing point
= freezing point of solution
= freezing point of water
i = Van't Hoff factor
= freezing point constant
m = molality
= mass of solute
= mass of solvent
= molar mass of solute
From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.
Hence, the molar mass of the solute would be low.
Answer : The correct option is (A).
Explanation :
- Endothermic reaction : When two liquids are combine into a flask, then the flask feels cold when we touch it because the system absorbed heat from the surrounding.
In general, endothermic process absorbs heat and cool the surrounding.
- Exothermic reaction : When two liquids are combine into a flask, then the flask feels hot when we touch it because the system released heat into the surrounding.
In general, exothermic process releases heat and rise the temperature of surrounding.
Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻
1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions
the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol
NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol
total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³