Answer:
In the option(A) moles of HCl left are 0.100 moles which is wrong, making the option incorrect.
Explanation:

Moles of HCl = n
Molarity of HCl = 1.0M
Volume of HCl solution = 30.0 mL = 0.030 L (1 mL = 0.001L)


Moles of Fe = 
According to recation , 1mol of Fe reacts with 2 mol HCl. Then 0.01 mole of Fe will recat with :
of HCl
This means that HCl uis in excess , hence excessive reagent.
Moles of HCl left unreacted :
= 0.030 mol - 0.020 mol = 0.010 mol
But in the option moles of HCl left are 0.100 moles which is wrong, making the option incorrect.
A cold front is the leading edge of a cooler mass of air, replacing at ground level a warmer mass of air, which lies within a fairly sharp surface trough of low pressure.
Answer:
Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)
Explanation:
Use of aqueous sodium hydroxide is a precipitation reaction to test for anions or cations. Aqueous sodium hydroxide in a precipitate test forms a insoluble precipitates along with some colors characteristics.
Aqueous sodium hydroxide (NaOH) when mixed with copper(II) (Cu2+) forms a blue precipitate. The formula is as follows:
Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)
Answer:
Volume of stock solution needed = 6.0299 mL
Explanation:
<u>
</u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.
This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.
<u>Data:</u>
M1 = 6.01 M stock solution concentration
M2 = 0.3624 M diluted solution concentration
V2 =100 mL diluted solution volume
V1 = ? stock solution volume
M1 * V1 = M2 * V2

<span>answer is A : attracted to the negative terminal of the voltage source
I think, that the "hole"moves as it captures a free electron leaving another hole in a slightly different place. The electron leaving leaves a net + charge, which is attracted to the negative terminal. Because the "hole" behaves as a positive charge it is attracted towards the negative terminal.</span>