Answer:
i need a picture to solve
Explanation:
It's 10.
Mass = density x volume
M = 1g/ml(10ml) = 10g
When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g
Answer:
MgCO₃
Explanation:
From the question given above, we obtained:
MgF₂ + Li₂CO₃ —> __ + 2LiF
The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:
MgF₂ (aq) —> Mg²⁺ + 2F¯
Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯
MgF₂ + Li₂CO₃ —>
Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:
MgF₂ + Li₂CO₃ —> __ + 2LiF
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Therefore, the missing part of the equation is MgCO₃
Answer:
The resulting structure is:
R-CH-CH2 (OH)-CH-CH2(OH)-CH-CH2(OH)-CH-...
Explanation:
It can be said that vinyl alcohol cannot be polymerized by addition polymerization, since a very unstable radical would be formed, which is not possible. Generally, it is formed by polymerization and subsequent hydrolysis of the vinyl polyacetate. According to the exercise, the structure that results with its four units can be seen that the hydroxyl groups are found in the methylene carbons
