Explanation:
1) refine the specimen into fine powder 2) place the smallest amount you can see in the capillary tube 3) set the voltage to increase exponentially to 200 below the predicted temperature, then adjust so that the temperature rises to 20 per minute 4) report the temperature at which the liquid first appears and the temperature at which the last crystal disappears.
Answer:
The empirical formula is SF6 (option E)
Explanation:
Step 1: Data given
Mass of sulfur = 3.21 grams
Mass of fluorine = 11.4 grams
Molar mass sulfur = 32.065 g/mol
Molar mass fluorine = 19.00 g/mol
Step 2: Calculate moles
Moles = mass /molar mass
Moles sulfur = 3.21 grams / 32.065 g/mol
Moles sulfur = 0.100 moles
Moles fluorine = 11.4 grams / 19.00 g/mol
Moles fluorine = 0.600 moles
Step 3: Calculate mol ratio
We divide by the smallest amount of moles
S: 0.100 / 0.100 = 1
F : 0.600 / 0.100 = 6
The empirical formula is SF6 (option E)
Answer: plants and animal cells
Explanation:
Answer:
Q = 60192 j
Explanation:
Given data:
Volume of water = 0.45 L
Initial temperature = 23°C
Final temperature = 55°C
Amount of heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 55°C - 23°C
ΔT = 32°C
one L = 1000 g
0.45 × 1000 = 450 g
Specific heat capacity of water is 4.18 j/g°C
Q = m.c. ΔT
Q = 450 g. 4.18 j/g°C. 32°C
Q = 60192 j