Answer:
The initial temperature of the metal is 84.149 °C.
Explanation:
The heat lost by the metal will be equivalent to the heat gain by the water.
- (msΔT)metal = (msΔT)water
-32.5 grams × 0.365 J/g°C × ΔT = 105.3 grams × 4.18 J/g °C × (17.3 -15.4)°C
-ΔT = 836.29/12.51 °C
-ΔT = 66.89 °C
-(T final - T initial) = 66.89 °C
T initial = 66.89 °C + T final
T initial = 66.89 °C + 17.3 °C
T initial = 84.149 °C.
Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
276 grams of carbon in 23.0 moles
OK in the case of hydrazine 14 grams of nitrogen combine with 2 gram of hydrogen and with ammonia 14 grams combine with 3 grams of hydrogen.
Ratio 2:3
