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:
14. 13.2cg = 1.32dg
15. 3.8m = 0.0038km
16. 24.8L = 24800mL
17. 0.87kL = 870L
18. 26.01cm = 0.0002601km
19. 0.001hm = 10cm
Explanation:
14. 13.2/10 = 1.32
15. 38/1000 = 0.0038
16. 24.8(1000) = 24,800
17. 0.87(1000) = 870
18. 26.01/100000 = 0.0002601
19. 0.001hm(10000) = 10
An easy way to do these by yourself is to familiarize yourself with what each prefix means. Once you do this, you can multiply the value of the prefix when converting from a smaller unit of measurement to a larger one and divide the value of the prefix when converting from a large unit of measurement to a smaller one.
First one is reactants and then second one is products
A substance that is impenetrable by x-rays is described as being radiopaque.
Radiopaque substances will not allow x-rays and or other forms of radiations to pass through them.
Instead, they absorb or block the rays and when used in radiology, they appear white or light gray on photographic films.
Radiopaque materials are applied in generating ultrasound images and other forms of clinical procedures.
More on radiopaque materials can be found here: brainly.com/question/10583205?referrer=searchResults
Answer:
H₂O is the limiting reactant
Theoretical yield of 240 g Al₂O₃ and 14 g H₂
Explanation:
Find how many moles of one reactant is needed to completely react with the other.
6.5 mol Al × (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O
We need 9.75 mol of H₂O to completely react with 6.5 mol of Al. But we only have 7.2 mol of H₂O. Therefore, H₂O is the limiting reactant.
Now find the theoretical yield:
7.2 mol H₂O × (1 mol Al₂O₃ / 3 mol H₂O) × (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃
7.2 mol H₂O × (3 mol H₂ / 3 mol H₂O) × (2 g H₂ / mol H₂) ≈ 14 g H₂
Since the data was given to two significant figures, we must round our answer to two significant figures as well.
