Answer:
This metal could be the aluminium with a specific heat of
Explanation:
A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows
We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that
Finally, the specific heat of this metal is
The aluminium could be the metal, its specific heat is similar to that found in this problem.
Finally, we can conclude that this metal could be the aluminium with a specific heat of
23.0 + 60.0 = 83.0° C heat energy is required to raise
The balanced chemical reaction would be
<span>fecl2 + 2naoh = fe(oh)2(s) + 2nacl
Initial amounts of the reactants are given, so, we need to determine which of the reactants is the limiting reactant and use this amount to determine what is asked. However, what is being asked is how many of the FeCl2 is used in the reaction, showing that it is NaOH that is the limiting reactants. Thus, we just use the initial amount of NaOH and relate the substances by the chemical reaction as follows:
6 mol NaOH ( 1 mol FeCl2 / 2 mol NaOH ) = 3 mol FeCl2
Therefore, 3 moles of FeCl2 is used up and 3 moles of FeCl2 is also left after the reaction.</span>
Answer:
Carbon 3 is double bonded to an oxygen and attached to carbon 2 and carbon 4. :
Answer: Carbonyl group ( Ketone or aldehyde)
Carbon 17 is attached to an oxygen, which is attached to a hydrogen. :
Answer: Carboxyl group (Carboxylic acid)
A central carbon is attached to an amine, two hydrogens, and a carbon that is double bonded to an oxygen and single bonded to an oxygen attached to a hydrogen. :
Answer: Amide group
An amide group contains both amine and carboxyl
Answer:
60 grams of ice will require 30.26 calories to raise the temperature 1°C.
Explanation:
The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released or absorbed by the system.
m is the mass of the ice (m = 60.0 g).
c is the specific heat capacity of ice (c = 2.108 J/g.°C).
ΔT is the temperature difference (ΔT = 1.0 °C).
∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.
<em>It is known that 1.0 cal = 4.18 J.</em>
<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>