Hey there!:
Write the molecular equation for the reaction of MgSO4 with Pb(NO3)2 :
MgSO4(aq) + Pb(NO3)2(aq) ---> Mg(NO3)2(aq) + PbSO4(s)
Write the total ionic equation for the reaction :
Mg²⁺ (aq) + SO₄⁻² (aq) + Pb²⁺ (aq) + 2 NO₃⁻¹ (aq) + PbSO₄(s)
Therefore:
Cancel the spectator ions on both sides:
Pb²⁺ (aq) + SO₄⁻² (aq) ---> PbSO4(s)
Hope that helps!
Answer:
- Rock composition
- Exposed surface area
Explanation:
Although there are other factors involved, I have selected two most important factors that could play a vital role in the resistance of the rock weathering. Here I have discussed each of them briefly.
<em>Rock composition:</em> If a rock is compact and composed of minerals which are not easy to disintegrate by physical as well as chemical weathering, it will increase the resistance of natural rock weathering process. For example, if a rock is composed of high silica content, it will not easy weathered as compared to the rock which is composed of feldspar, calcite, and/or iron. Likewise, igneous rocks are more resistant to the sedimentary rocks because the minerals composition is more compact and dense.
<em>Exposed surface area: </em>If a rock is exposed to the natural environment, it could be weathered more quickly then the rock which is not exposed directly (e.g. underlying rocks). Therefore, covering a rock surface or reducing the surface area could contribute towards increasing resistance of rock weathering. This is the reason that deep underlying rocks stay intact for a longer period of time as compared the those which are exposed.
For this problem, we use the freezing point depression formula:
Tf,solvent - Tf,solution = Kf×m
Where
Tf,solvent is freezing pt of the solvent
Tf,solution is freezing pt of solution
Kf is the <span>molal freezing point depression constant of the solvent
m is the molality equal to mol solute/kg solvent
For water as solvent, Tf,solvent = 0</span>°C; Kf = 1.86 °C/m:
0°C - ⁻10.5 °C = (1.86 °C/m)(m)
Solving for m,
m = 5.645 mol solute/kg solvent
5.645 = mol solute/1 kg water
mol solute = 5.645 mol fructose
Since the molar massof fructose is 180.16 g/mol,
Mass of Fructose = 5.645 mol * 180.16 g/mol
Mass of Fructose = 1,017.03 g fructose or 1.017 kg fructose
K = 39.09
Mn = 54.93
O = 15.99
KMnO4 = 39.09 + 54.93 + 15.99 x 4 => 157.98 g/mol
hope this helps!