Answer:
The molality is 1.15 m.
Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.
Calculate moles of H₂SO₄ from molarity:
C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles
Mass of solvent (water) based on density:
m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg
Therefore, molality is:
m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m
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According to Arrhenius, bases are defined as substances that produce OH⁻ ions when they dissolve in water. The anion NO₂⁻ fits this definition as illustrated in the following reaction:
NO₂⁻(aq) + H₂O (l) HNO₂ (aq) + OH⁻(aq)
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Answer:
In blood: Dispersed phase: blood cells; Dispersed medium: liquid plasma
In fruit jelly: Dispersed phase: fruit juice; Dispersed medium: pectin
Explanation:
The dispersed phase refers to the phase where colloidal particles are dispersed within another phase, known as the dispersion medium.
In blood, the tiny cells act as colloidal particles, forming the dispersed phase within the liquid medium identified as plasma.
Conversely, in fruit jelly, the fruit juice constitutes the dispersed phase while the solid pectin serves as the dispersed medium.
Answer:
The ratios arranged in ascending order are; The ratio of the mass of Y to X in XY2 divided by the mass of Y to X in XY, The ratio of the mass of Y to X in XY3 divided by the mass of Y to X in XY, The ratio of the mass of Y to X in XY4 divided by the mass of Y to X in XY
1) Mass ratio = 3
2) Mass ratio = 2
3) Mass ratio = 4
Explanation:
Comprehensive calculations are displayed in the attachment.
<span>(NH4)2CO3 -> 96.09 g/mol
(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate
In this calculation, the unit 'grams' cancels out as it's present in both the numerator and the denominator, leading to 'mol' being the remaining unit.
Examining the formula (NH4)2CO3, it can be interpreted as:
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3
This means every mole of ammonium carbonate yields one mole of carbonate ions and two moles of ammonium ions.
(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion) + (0.363981 mol ammonium ion) </span>
