Answer:The surface heated air expands as it warms, becomes less dense than surrounding cooler air and rises as buoyant and turbulent bubbles. This is convection and is the main process by which the troposphere mixes and heats. Although convection stirs and mixes the troposphere, the higher it is the colder it becomes.
Explanation: Read this and you'll get your answer~! I hope i helped you~! Have an GREAT day too~! <\3
First, find moles of oxygen gas: (3.01 x10^23 molec.)/(6.02 x10^23) =0.5mol O2
Second, multiply moles by the standard molar volume of a gas at STP:(0.5mol)(22.4L) = 11.2L O2
Answer: v2=331.289mL
Explanation:
Formula for ideal gas law is p1v1/T1=p2v2/T2
P1=782.3mmHg
P2=769mmHg at STP
V1=362.4mL
V2=?
T1=273+34.4=307.4k
T2=273k at STP
Then apply the formula and make v2 the subject of formula
V2= 782.3×362.4×273/760×307.4
V2=77397006.96/233624
V2=331.289mL
Answer:
Decreases
Explanation:
F = GM1M2/R²
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Answer:
54g of water
Explanation:
Based on the reaction, 1 mole of methane produce 2 moles of water.
To solve this question we must find the molar mass of methane in order to find the moles of methane added. With the moles of methane and the chemical equation we can find the moles of water produced and its mass:
<em>Molar mass CH₄:</em>
1C = 12g/mol*1
4H = 1g/mol*4
12g/mol + 4g/mol = 16g/mol
<em>Moles methane: </em>
24g CH₄ * (1mol / 16g) = 1.5 moles methane
<em>Moles water:</em>
1.5moles CH₄ * (2mol H₂O / 1mol CH₄) = 3.0moles H₂O
<em>Molar mass water:</em>
2H = 1g/mol*2
1O = 16g/mol*1
2g/mol + 16g/mol = 18g/mol
<em>Mass water:</em>
3.0moles H₂O * (18g / mol) =
<h3>54g of water</h3>
