The domain that is an onion is the <span>Eukaryote.
</span>
Observation/ question
research
hypothesis
<span>experiment
</span>analysis
conclusion
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:
2 KClO3 (s) = 2 KCl (s) + 3 O2 (g)
2.5 g x g
Explanation:
x g O2 = 2.5 g KClO3 x (1 mol KClO3) x (3 mol O2) x (32 g O2) = 0.98 g O2
(122.5 g KClO3) (2 mol KClO3) (1 mol O2)
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
2.5 g x g
x g KCl = 2.5 g KClO3 x (1 mol KClO3) x (2 mol KClO3) x (74.5 g KCl) = 1.52 g KCl
(122.5 g KClO3) (2 mol KClO3) (1 mol KCl)
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)
x mol 10 mol
x mol KClO3 = 10 mol O2 x (2 mol KClO3) = 6.7 mol KClO3
(3 mol O2)
Answer:
It might be responding variable.
