Answer:
0.70 J/g.°C
Explanation:
Step 1: Given data
- Mass of graphite (m): 402 g
- Heat absorbed (Q): 1136 J
- Initial temperature: 26°C
- Specific heat of graphite (c): ?
Step 2: Calculate the specific heat of graphite
We will use the following expression.
Q = c × m × ΔT
c = Q / m × ΔT
c = 1136 J / 402 g × (30°C - 26°C)
c = 0.70 J/g.°C
Answer:
Explanation:
Ok so an atom is each ball. So in the first one there are 5 balls. In the second one there are 4 and so on. A molecule contains more than two balls. So they are all molecules. For the counting reactants and products, count how many balls are to the left of the arrow which is your number of reactants and count the balls to the right to find the number of product atoms.
The correct answer is "B. Make a Hypothesis".
We can use the ideal gas equation:
PV = nRT
P = 202.6kPa = 202600 Pa (You have to
multiply by 1000)
n = 0.050 mole
R = 0.082 atm*l/(K*mol)
T = 400K
We will have to convert from Pa to atm or
viceversa.
101325 Pa________1 atm
202600 Pa________x = 2.00 atm
2atm*V = 0.050 mole*0.082 atm*l/(K*mol)* 400K
V = 0.050 mole*0.082 atm*l/(K*mol)* 400K/2atm
= 0.82 liters = 820 mililiters
