Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
If the refrigerator has no room to cool the meat that you've just cooked, then you should clear it with other stuff or food that has not been consumed for days. There might be some goods that are not good for consumption anymore even though it has been placed in the ref for a couple of days. Hope this answers your question.
Explanation:
<h3 /><h2>
<em><u>H2 </u></em><em><u>+</u></em><em><u> </u></em><em><u>O2 </u></em><em><u>=</u></em><em><u> </u></em><em><u>H2O</u></em></h2>
<h2>
<em><u>Hydrogen</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>Oxygen</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>Water</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em></h2>
<em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em>
Answer:
88,88 % de O y 11,11 % de H
Explanation:
La composición porcentual se define como la masa que hay de cada mol de átomo en 100g. Las moles de agua en 100g son:
<em>Masa molar agua:</em>
2H = 2*1g/mol = 2g/mol
1O = 1*16g/mol = 16g/mol
Masa molar = 2 + 16 = 18g/mol
100g H2O * (1mol / 18g) = 5.556 moles H2O.
Moles de hidrógeno:
5.556 moles H2O * (2mol H / 1mol H2O) = 11.11 moles H
Moles Oxígeno = Moles H2O = 5.556 moles
La masa de hidrógeno es:
11.11mol * (1g/mol) 11.11g H
La masa de oxígeno es:
5.556 mol * (16g / 1mol) = 88.89g O
Así, el porcentaje de O es 88.89% y el de H es 11.11%. La opción correcta es:
<h3>88,88 % de O y 11,11 % de H</h3>
