Answer: An amount of
heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.
Explanation:
Given: mass of lead = 4.64 kg
Convert kg into grams as follows.



The standard value of specific heat of lead is
.
Formula used to calculate heat is as follows.

where,
q = heat energy
m = mass of substance
C = specific heat of substance
= change in temperature
Substitute the value into above formula as follows.

Thus, we can conclude that
heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.
Answer:
The final temperature at 1050 mmHg is 134.57
or 407.57 Kelvin.
Explanation:
Initial temperature = T = 55
= 328 K
Initial pressure = P = 845 mmHg
Assuming final to be temperature to be T' Kelvin
Final Pressure = P' = 1050 mmHg
The final temperature is obtained by following relation at constant volume

The final temperature is 407.57 K
Al(s) + 2AgNO3(aq) = Al(NO3)3(aq) + 3 Ag (s)
is a single replacement reaction
A single replacement reaction is a type of chemical reaction were element react with a compound and the element take place of another element in that compound. In the reaction above Aluminium (Al) take the place of silver (Ag) from it compound.that is AgNO3. single replacement reaction is possible because aluminium is in high in reactivity series as compared to silver.
Volume of base(NaOH)= Vb =5 mL
Concentration of base(NaOH)= Mb =2 M
Volume of acid(HCl)= Va =10 mL
Concentration of acid(HCl)= Ma = ?
Now,
Ma*Va = Mb*Vb
Ma*(10) = (2)*(5)
Ma= 1M