Answer:
its B
Explanation:
Because its can only go cold to hot. -hope this help :)
Answer:
The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Explanation:
We are given that
Aqueous solution that contains 22.9% NaOH by mass means
22.9 g NaOH in 100 g solution.
Mass of NaOH(WB)=22.9 g
Mass of water =100-22.9=77.1
Na=23
O=16
H=1.01
Molar mass of NaOH(MB)=23+16+1.01=40.01
Number of moles =
Using the formula
Number of moles of NaOH

Molar mass of water=16+2(1.01)=18.02g
Number of moles of water

Now, mole fraction of NaOH
=

=0.882
Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Answer:
1. 0.97 V
2. 
Explanation:
In this case, we can start with the <u>half-reactions</u>:


With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

I hope it helps!
Answer:
Explanation:
phosphorus belongs to group 5 of the periodic table because it has 5 electron in its outermost shell the number of electron in the outermost shell of electron determine the group of the element in the periodic table
The mass of lime that can be produced from 4.510 Kg of limestone is calculated as below
calculate the moles of CaCO3 used
that is moles =mass/molar mass
convert Kg to g = 4.510 x1000 =4510g
= 4510 / 100 =45.10 moles
CaCO3 = CaO +O2
by use of mole ratio between CaCO3 to CaO (1:1) the moles of CaO is also= 45.10 moles
mass of CaO = moles x molar mass
45.10 x56 = 2525.6 g of CaO