Matter can be classified according to physical and chemical properties. Matter is anything that occupies space and has mass. The three states of matter are solid, liquid, and gas. Extensive properties depend on the amount of material and include mass and volume.
The answer is B.
Explanation:
The other options contain K, L, or Mg which are all metals.
Answer:
The enthalpy change during the reaction is -7020.09 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
where,
q = heat gained = ?
c = specific heat =
= final temperature =
= initial temperature =
Now put all the given values in the above formula, we get:
The heat gained by water present in calorimeter. = q'
where,
q' = heat gained = ?
m = mass of water =
c' = specific heat of water =
= final temperature =
= initial temperature =
q ' = 12,060.38 J
Now we have to calculate the enthalpy change during the reaction.
where,
= enthalpy change = ?
Q = heat gained = -(q+q') = -(2,485.25 J + 12,060.38 J)= -14,545.63 J
Q = -14.54563 kJ
n = number of moles fructose =
Therefore, the enthalpy change during the reaction is -7020.09 kJ/mole.
Non metals.
metalloids are substances that have both properties of metallic and non-metallic
Answer:
The final temperature of the mixture is 22.3°C
Explanation:
Assuming that the 120 g substance at 80°C is water, final temperature of the mixture can be determined using the formula:
Heat lost = Heat gained
Heat = mc∆T where m is mass, c is specific heat capacity of water, and ∆T is the temperature change =<em> Tfinal - Tinitial</em>.
Let the final temperature be T
Heat lost = 120 × c × (T - 80)
Heat gained = 3000 × c × ( T - 20)
Equating the heat lost and heat gained
120 × c × -(T - 80) = 3000 × c × (T - 20)
9600 - 120T = 3000T - 60000
60000 + 9600 = 3000T + 120T
69600 = 3120T
T = 69600/3120
T = 22.3°C
Therefore, the final temperature of the mixture is 22.3°C