The thing that two changes have in common that snails growing shells and rust forming on a bicycle frame is option D. Both are caused by cooling.
<h3>How come snails develop shells?</h3>
Calcium carbonate is said to be the material that makes up the shell. The snail's shell expands as it grows to accommodate its growing body. Snails and slugs are also members of the mollusc family of creatures.
Therefore, note that air that has been mixed with the metal can make rust to develop. and as such, option D. Both are caused by cooling. is correct.
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What do these two changes have in common? snails growing shells rust forming on a bicycle frame Select all that apply.
A. Both are only physical changes.
B. Both are caused by heating.
C. Both are chemical changes.
D. Both are caused by cooling
This equation C5H + O2 ---> CO2 + H2O has a mistake.
C5H is wrong. You missed the subscript of H.
I will do it for you assuming some subscript to show you the procedure, but you have to use the right equation to get the right balanced equation.
Assuming the tha combustion equation is C5H12 + O2 ---> CO2 + H2O
First you need to balance C, so you put a 5 before CO2 and get
C5H12 + O2 ---> 5CO2 + H2O
Now you count the hydrogens: 12 on the left and 2 on the right. So put a 6 before H2O and get:
C5H12 + O2 ---> 5CO2 + 6H2O
Now count the oxygens: 2 on the left and 16 on the right, so put an 8 on before O2:
=> C5H12 + 8O2 ---> 5CO2 + 6H2O.
You can verify that the equation is balanced
12.5g, each 10 years you lose a half of what you have at that given moment
Answer:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
Answer:
The amount of NO₂ that can be produced 8.533 g
Explanation:
According to question
2 NO(g) + O₂(g) → 2 NO₂(g)
Given
Moles of nitrogen monoxide = 0.377
Moles of oxygen = 0.278
Since 'NO' is the limiting reagent according to this ratio.
According to equation
2 moles NO reacts to form 2 moles NO₂
So, 0.1855 moles NO give = 0.1855 moles of NO₂
Mass of 1 mole NO₂ = 46 g/mole
Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g
