Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.
Explanation:
Given: 
= 61 L, 
= 183 K, 
= 0.60 atm
At STP, the value of pressure is 1 atm and temperature is 273.15 K.
Now, formula used to calculate the new volume is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.
I think the answer is 2Mg + H2O4 = Cu12O4 + 2H
I’m really not sure though so it might be wrong… I’m not the best at balancing equations lol
Answer:
hola soy jess, tu respuesta esta aqui
¿cuantos moles de CO2 se requiere para reaccionar 2 moles de Ba(OH)2
2 mol Ba(OH)₂ × \frac{1molCO_{2} }{1molBa (OH)_{2}}
1molBa(OH)
2
1molCO
2
= 2 moles CO₂
Explanation:
espero que pueda ayudarte
hermana/hermano
lo que
hahahaha
Answer:
32.7 g of Zn
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
Zn + 2HCl —> ZnCl₂ + H₂
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂
Next, we shall determine the number of mole of Zn required to produce 0.5 mole of H₂. This can be obtained as follow:
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂.
Therefore, 0.5 mole of Zn will also react to produce to 0.5 mole of H₂.
Thus, 0.5 mole of Zn is required.
Finally, we shall determine the mass of 0.5 mole of Zn. This can be obtained as follow:
Mole of Zn = 0.5 mole
Molar mass of Zn = 65.4 g/mol
Mass of Zn =?
Mass = mole × molar mass
Mass of Zn = 0.5 × 65.4
Mass of Zn = 32.7 g
Thus, 32.7 g of Zn is required to produce 0.5 mole of H₂.
Respiring while swimming underwater
Because anaerobic respiration means respiration with no oxygen, and there is no oxygen underwater
