Answer:
27.60 g urea
Explanation:
The <em>freezing-point depression</em> is expressed by the formula:
In this case,
- ΔT = 5.6 - (-0.9) = 6.5 °C
m is the molality of the urea solution in X (mol urea/kg of X)
First we<u> calculate the molality</u>:
- 6.5 °C = 7.78 °C kg·mol⁻¹ * m
Now we<u> calculate the moles of ure</u>a that were dissolved:
550 g X ⇒ 550 / 1000 = 0.550 kg X
- 0.84 m = mol Urea / 0.550 kg X
Finally we <u>calculate the mass of urea</u>, using its molecular weight:
- 0.46 mol * 60.06 g/mol = 27.60 g urea
Answer:
Net ionic equation:
Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)
Explanation:
Chemical equation:
ZnCl₂ + KOH → KCl + Zn(OH)₂
Balanced chemical equation:
ZnCl₂ + 2KOH → 2KCl +Zn(OH)₂
Ionic equation;
Zn²⁺(aq) + 2Cl⁻(aq) + 2K⁺(aq) + 2OH⁻(aq) → 2K⁺(aq) + 2Cl⁻(aq) +Zn(OH)₂(s)
Net ionic equation:
Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)
The K⁺ and Cl⁻ are spectator ions that's why these are not written in net ionic equation. The Zn(OH)₂ can not be splitted into ions because it is present in solid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
Answer:
<u>Our beaches would be unprotected</u>
In the short-term, these artificial sand hills will be destroyed by the elements. Because sand dunes protect inland areas from swells, tides, and winds, they must be protected and defended like national treasures. ... The ocean and the wind can have an unpredictable, destructive force on coastal regions.
- surfertoday
Natural sand dunes play a vital role in protecting our beaches, coastline and coastal developments from coastal hazards such as erosion, coastal flooding and storm damage. Sand dunes protect our shorelines from coastal erosion and provide shelter from the wind and sea spray.
- Waikato Regional Council
n = m / M
Where, n is moles of the compound (mol), m is the mass of the compound (g) and M is the molar mass of the compound (g/mol)
Here, the given ethanol mass = 50.0 kg = 50.0 x 10³ g
Molar mass of the ethanol = (12 x 2 + 1x 6 + 1 x 16) g/mol
= 46 g/mol
Hence, moles in 50.0kg of ethanol = 50.0 x 10³ g / 46 g/mol
= 1086.96 mol
