Answer:
The specific heat of gold is 0.129 J/g°C
Explanation:
Step 1: Data given
Mass of gold = 15.3 grams
Heat absorbed = 87.2 J
Initial temperature = 35.0 °C
Final temperature = 79.2 °C
Step 2:
Q = m*c*ΔT
⇒ Q =the heat absorbed = 87.2 J
⇒ m = the mass of gold = 15.3 grams
⇒ c = the specific heat of gold = TO BE DETERMINED
⇒ ΔT = The change in temperature = T2 - T1 = 79.2 - 35.0 = 44.2 °C
87.2 J = 15.3g * c * 44.2°C
c = 87.2 / (15.3 * 44.2)
c = 0.129 J/g°C
The specific heat of gold is 0.129 J/g°C
An occluded front is well known to bring a large amount of rain and snow (Option F).
<h3>What is an occluded front?</h3>
An occluded front is a mass of air capable of bringing precipitation in the form of liquid water (rain) or snow.
This meteorological phenomenon (occluded front) is formed by warm air trapped with cold air.
In conclusion, an occluded front is well known to bring a large amount of rain and snow (Option F).
Learn more about occluded front here:
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Answer:
The 2292 moles of CO are needed to react completely with 122 Kg of Fe₂O₃.
Explanation:
Given data:
Mass of Fe₂O₃ = 122 Kg ( 122×1000 = 122000 g)
Moles of CO = ?
Solution:
Chemical equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Number of moles of Fe₂O₃:
Number of moles = mass/ molar mass
Number of moles = 122000 g /159.69 g/mol
Number of moles = 764 mol
Now we will compare the moles of Fe₂O₃ with CO.
Fe₂O₃ : CO
1 : 3
764 : 3×764 =2292 mol
The 2292 moles of CO are needed to react completely with 122 Kg of Fe₂O₃.
3.81 kpa is the condition which is not true at STP
According to IUPAC the standard temperature and pressure that is STP the temperature is 273.15 k or 0 degrees celsius . and the absolute temperature of 101.325 Kpa or 1 atm. In addition at STP the volume of ideal gas is 22.4
Answer:Label the parts of this wave.
A:
✔ crest
B:
✔ amplitude
C:
✔ trough
D:
✔ wavelength
Explanation: