Answer:
0.862 J/gºC
Explanation:
The following data were obtained from the question:
Mass of metal (Mₘ) = 50 g
Initial temperature of metal (Tₘ) = 100 °C
Mass of water (Mᵥᵥ) = 400 g
Initial temperature of water (Tᵥᵥ) = 20 °C
Equilibrium temperature (Tₑ) = 22 °C
Specific heat capacity of water (Cᵥᵥ) = 4.2 J/gºC
Specific heat capacity of metal (Cₘ) =?
The specific heat capacity of the metal can be obtained as follow:
Heat lost by metal = MₘCₘ(Tₘ – Tₑ)
= 50 × Cₘ × (100 – 22)
= 50 × Cₘ × 78
= 3900 × Cₘ
Heat gained by water = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
= 400 × 4.2 × (22 – 20)
= 400 × 4.2 × 2
= 3360 J
Heat lost by metal = Heat gained by water
3900 × Cₘ = 3360
Divide both side by 3900
Cₘ = 3360 / 3900
Cₘ = 0.862 J/gºC
Therefore, the specific heat capacity of the metal is 0.862 J/gºC
Metals don't form covalent bonds because of the low ionization energes of the metal atoms. It is easier for them to release electrons rather than sharing it. But this is not always the case, there are some metals that can form covalent bonds.
Answer:
metal : Mercury(Hg)
non metal : bromine (Br)
Explanation:
mercury is liquid at room temperature and pressure and the same as bromine
Answer:
Relation between , molality and temperature is as follows.
T =
It is also known as depression between freezing point where, i is the Van't Hoff factor.
Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.
i for = 3
i for glucose = 1
i for NaCl = 2
Depression in freezing point will have a negative sign. Therefore, d
depression in freezing point for the given species is as follows.
=
=
=
Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.
Glucose < NaCl <
Explanation:
A mole of any gas occupied 22.4 L at STP. So, the number of moles of nitrogen gas at STP in 846 L would be 846/22.4 = 37.8 moles of nitrogen gas.
Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.
