You would also need to know the mol of sodium. After knowing the mol, write down the reaction;
Na + H20 -> NaOH + H2
Use the mol ratio, to find the amount of mol of sodium hydroxide. (Remember to balance the equation first)
After you have the necessary variables, use the following formula:
mass of sodium hydroxide = mol of sodium hydroxide x molar mass (relative/formula mass of sodium hydroxide)
That is how you will find the mass of sodium hydroxide.
Answer:
If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
In this way, between heat and temperature there is a direct proportional relationship (Two magnitudes are directly proportional when there is a constant so that when one of the magnitudes increases, the other also increases; and the same happens when either of the two decreases .). The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature, ΔT= Tfinal - Tinitial
In this case:
- Q= 27 kJ= 27,000 J (being 1 kJ=1,000 J)

- m=700 g
- ΔT= Tfinal - Tinitial= Tfinal - 90 °C
Replacing:

Solving:


16.125 °C= Tfinal - 90 °C
Tfinal= 16.125 °C + 90 °C
Tfinal= 106.125 °C
<u><em>If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C</em></u>
Answer:
Reduction
Explanation:
The charge on Mg goes from 2+ to 0 which means this is a reduction.
C. Aluminum (Al) oxidized, zinc (Zn) reduced
<h3>Further explanation</h3>
Given
Metals that undergo oxidation and reduction
Required
A galvanic cell
Solution
The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.

or:
E ° cell = E ° reduction-E ° oxidation
For the reaction to occur spontaneously (so that it E cell is positive), the E° anode must be less than the E°cathode
If we look at the voltaic series:
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)
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From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.
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