Answer:
the initial temperature of the iron sample is Ti = 90,36 °C
Explanation:
Assuming the calorimeter has no heat loss to the surroundings:
Q w + Q iron = 0
Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron ( final temperature of water= final temperature of iron = T )
Assuming Q= m*c*( T- Tir)
mc*cc*(T-Tc)+mir*cir*(T - Tir) = 0
Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )
Tir = 90.36 °C
Note :
- The specific heat capacity of water is assumed 1 cal/g°C = 4.186 J/g°C
- We assume no reaction between iron and water
Answer:
The new volume of the balloon is 38.5 L
Explanation:
Step 1: Data given
Volume at the start = V1 = 35.0 L
Temperature at the start = T1 = 303 Kelvin
Volume by 3pm = TO BE DETERMINED
Temperature by 3pm = 333 Kelvin
<u>Step 2: </u>Calculate the new volume
Charles' gas law says
V1/T1 = V2/T2
V
1 is the initial volume and T1 is the initial temperature
V2 is the final volume and T2 is the final temperature
35 L / 303 Kelvin = V2 / 333 Kelvin
V2 = 35L * 333 Kelvin / 303 Kelvin
V2 = 38.47L ≈ 38.5 L
The new volume of the balloon is 38.5 L
Answer:
A cuz a heterogeneous mixture is no uniform
Answer:
46.3g H2O
Explanation:
start by balancing it: CaC2(s) + 2H2O(g) -> Ca(OH)2(s) + C2H2(g)
then use factor label method to solve
82.4g CaC2 x (1 mol CaC2/64.10g CaC2) x (2 mol H2O/1 mol CaC2) x (18.016g H2O/1 mol H20) = 46.3g H2O
4 In the open chain, 5 in the cyclic. Just like glucose.
