<span>A physical change basically involves a change in physical properties. Some examples of physical properties include: texture, shape, size, color, volume, mass, weight, etc.
The water melting has changed the shape and therefore it's physical properties, but the chemical nature of the water has not been altered.
So that's why it's a physical change, and not a chemical change.</span>
Answer:
Explanation:
Hello there!
In this case, according to the given question about stoichiometry, it is possible for us to calculate the required moles of water that will be produced by 12 grams of hydrogen, by using the molar mass of this reactant (2.02 g/mol as it is diatomic) and the 2:2 mole ratio in the chemical equation by solving the following setup:
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Answer:
52
Explanation:
This is because if R is the midpoint of FRG, FR is half of FRG, so basically all you do it multiply by 2 to get the FRG
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
Answer:
Normalidad = 4N
%p/V = 27.6%
Explanation:
La solución 2M de carbonato de potasio contiene 2moles de carbonato por litro de solución. La normalidad son los equivalente de carbonato de potasio (2eq/mol) por litro de solución:
2moles * (2eq/mol) = 4eq / 1L = 4N
El porcentaje peso volumen es el peso de carbonato en gramos dividido en el volumen en mL por 100:
%p/V:
Masa K2CO3 -Masa molar: 138.205g/mol-
2moles * (138.205g/mol) = 276g K2CO3
Volumen:
1L * (1000mL/1L) = 1000mL
%p/V:
276g K2CO3 / 1000mL * 100
<h3>%p/V = 27.6%</h3>
