How many grams of lead (II) carbonate are produced when 170 grams of sodium nitrate are also produced by the following unbalanced equation? Pb(NO3)2 + Na2CO3 → PbCO3 + NaNO3

We need 0.482 L of kerosene

Explanation:

Step 1: Data given

Mass of the block tin = 1.59 kg

Initial temperature tin= 88.0 °C

Final temperature = 49.0 °C

Initial temperature of kerosene = 32.0 °C

Density of kerosene =  820 kg/m³

Specific heat of tin is 218 J/(kg · °C)

Step 2: Calculate mass of kerosene

Heat lost = heat gained

Qtin = -Qkerosene

Q = m*c*ΔT

m(tin) * c(tin) * ΔT(tin) = -m(kerosene) * c(kerosene) * ΔT(kerosene)

⇒ mass of tine = 1.59 kg

⇒ c(tin) = the specific heat of tin = 218 J/ kg*°C

⇒ ΔT(tin) = The change in temperature = T2 - T1 = 49.0 °C - 88.0 °C = -39.0 °C

⇒ mass of kerosene = TO BE DETERMINED

⇒ c(kerosene) = The specific heat of kerosene =  2010 J/kg*°C

⇒ ΔT = 49.0 - 32.0 = 17.0

1.59 kg * 218 J/kg*°C * -39.0 °C = - m(kerosene) * 2010 J/kg*°C *17.0 °C

-13518.18 = -m(kerosene) * 34170

m(kerosene) = 0.39562 kg

Step 3: Calculate volume of kerosene

Volume = mass / density

Volume = 0.39562 kg / 820 kg/m³

Volume = 4.82 * 10^-4 m³ = 0.482 L

We need 0.482 L of kerosene