Answer:
C)There would have been four pots in both cases
Explanation:
In an incomplete reaction of naphthalene and 2-bromo-2-methylpropane, we started with 2 spots (one each for the two starting materials) and we should end up with 3 spots If spots will appear clearly, one for naphthalene, one for the alkyl halide and one for the product. However, this is possible to have same number of spots initially and finally when we consider 2 spots each for naphthalene and 2-bromo-2-methylpropane then after reaction we have 2 spots of products as each 1 spot of reactants decreases. Finally making initial 4 and final 2+1+1 =4.
So, the answer is (C)
Answer: mole fraction of methanol = 0.590
mole fraction of ethanol = 0.410
Explanation:
We are given:
Equal masses of methanol 
and ethanol 
are mixed.
let the mass be x g.
Calculating the moles of methanol in the solution, by using the equation:
Calculating the moles of ethanol in the solution, by using the equation:
To calculate the mole fraction of methanol, we use the equation:
To calculate the mole fraction of ethanol, we use the equation:
Thus mole fraction of methanol is 0.590 and mole fraction of ethanol 0.410 in three significant figures.
Answer:
A cell's plasma membrane, also known as the cell membrane, provides protection. It also maintains a constant atmosphere within the cell, and the membrane serves a variety of purposes. The first is to transport nutrients into the cell, and the second is to transport toxic compounds out
Explanation:
Answer:
The 3p orbitals have the same general shape and are larger than 2p orbitals, but they differ in the number of nodes. You have probably noticed that the total number of nodes in an orbital is equal to n−1 , where n is the principal quantum number. Thus, a 2p orbital has 1 node, and a 3p orbital has 2 nodes.
Answer:
the water concentration at equilibrium is
⇒ [ H2O(g) ] = 0.0510 mol/L
Explanation:
- CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)
∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30
⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L
⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
replacing in Kc:
⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30
⇒ 0.0721 [ H2O(g) ] = 3.679 E-3
⇒ [ H2O(g) ] = 0.0510 mol/L
