Answer: -112200J
Explanation:
The amount of heat (Q) released from an heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = ?
Mass of water vapour = 30.0g
C = 187 J/ G°C
Φ = (Final temperature - Initial temperature)
= 100°C - 120°C = -20°C
Then apply the formula, Q = MCΦ
Q = 30.0g x 187 J/ G°C x -20°C
Q = -112200J (The negative sign does indicates that heat was released to the surroundings)
Thus, -112200 joules of heat is released when cooling the superheated vapour.
Answer:
24.2 mL.
Explanation:
<em>Assuming constant temperature</em>, we can solve this problem using <em>Boyle's law</em>, which states:
Where:
We <u>input the data</u>:
- 0.98 bar * 25 mL = 1.013 bar * V₂
And <u>solve for V₂</u>:
The closest option is the second one: 24.2 mL.
The name of Pb(NO3)2 of is Lead II trioxonitrate V
The bond which exist in is Lead II trioxonitrate V ( Pb(NO3)2 ) are both ionic and covalent bond
<h3>What is a compound?</h3>
A compound is a substance which contains two or more elements chemically combined together
So therefore, the name of Pb(NO3)2 of is Lead II trioxonitrate V
Learn more about a chemical compound:
brainly.com/question/26487468
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6.069 grams is the mass of hydrogen formed when 27 g of aluminum reacts with excess hydrochloric acid.
Explanation:
Balanced equation for the reaction:
2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
data given:
mass of aluminum = 27 grams
atomic mass of one mole of aluminum = 26.89 grams/mole
formula to calculate number of moles:
number of moles = 
number of moles = 
= 1.004 moles of aluminum will react
from the balanced equation:
2 moles of Al reacted to form 3 moles of H2
1.004 moles of Al will produce x moles of H2
= 
x = 3.012 moles of H2 will be formed.
mass will be calculated as number of moles multiplied by atomic weight
mass of 3.012 moles of hydrogen ?(atomic weight of one mole H2 = 2.015 grams)
= 3.012 x 2.015
= 6.069 grams of H2 will be formed.
Answer:
7.5 mol
Step-by-step explanation:
2Al + 3Br₂ ⟶ 2AlBr₃
You want to convert moles of AlBr₃ to moles of Br₂.
The molar ratio is 3 mol Br₂:2 mol AlBr₃.
Moles of Br₂ = 5 × 3/2
Moles of Br₂ = 7.5 mol Br₂
You need 7.5 mol of Br₂ to produce 5 mol of AlBr₃.
