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FinnZ [79.3K]
3 years ago
10

The stability of a nucleus is determined by the ratio of electrons to protons.

Chemistry
1 answer:
Lera25 [3.4K]3 years ago
5 0
B. False
Stability is determined by the ratio of neutrons and protons. Electrons are not in nucleus.
You might be interested in
Chemistry. Will mark Brainly.
kogti [31]

Answer:

c is right

Explanation:

8 0
2 years ago
If 61.5 moles of an ideal gas occupies 97.5 liters at 473 K, what is the pressure of the gas, in mmHg?
kiruha [24]
I’d say for the answer 13.13 mmHg?
5 0
2 years ago
the combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110kg of carbon dioxide what is the limittin
mote1985 [20]

Answer:

Propane

Explanation:

From the question given, we were told that 0.1240 kg of propane reacted with excess oxygen to produce 0.3110kg of carbon dioxide.

Since the reaction took place in the presence of excess oxygen, therefore, propane is the limiting reactant as all of it is used up in the presence of excess oxygen.

4 0
3 years ago
Read 2 more answers
The gauge pressure inside a vessel is ‐40kPa, at an elevation of 5000m. a) What is the absolute pressure? b) At this elevation,
hoa [83]

Answer:

a) Pabs = 48960 KPa

b) T = 433.332 °C

Explanation:

  • Pabs = Pgauge + d*g* h

∴ d = 1000 Kg/m³

∴ g = 9.8 m/s²

∴ h = 5000 m

∴ P gauge = - 40 KPa * ( 1000 Pa / KPa ) = - 40000 Pa; Pa≡Kg/m*s²

⇒ Pabs = - 40000 Kg/ms² + ( 1000 Kg/m³ * 9.8 m/s² * 5000 m )

⇒ Pabs = 48960000 Pa = 48960 KPa

a) at that height and pressure, we find the temperature at which the water boils by means of an almost-exponential graph which has the following equation:

P(T) = 0.61094 exp ( 17.625*T / ( T + 243.04 ))......P (KPa) ∧ T (°C)....from literature

∴ P = 48960 KPa

⇒ ( 48960 KPa / 0.61094 ) = exp ( 17.625T / (T+ 243.04))

⇒ 80138.803 = exp ( 17.625T / ( T + 243.04))

⇒ Ln ( 80138.803) = 17.625T / ( T + 243.04))

⇒ 11.292 * ( T + 243.04 ) = 17.625T

⇒ 11.292T + 2744.289 = 17.625T

⇒ 2744.289 = 17.625T - 11.292T

⇒ 2744.289 = 6.333T

⇒ T = 433.332 °C

3 0
2 years ago
Cuando se quema 1 mol de metano –o sea, 16 g–, se desprenden 802
Anvisha [2.4K]

Answer:

1 gramo de metano aporta 50.125 kilojoules.

1 gramo de metano aporta 48.246 kilojoules.

Explanation:

La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (Q), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (\bar {Q}), en kilojoules por mol, dividido por su masa molar (M), en gramos por mol:

Q = \frac{\bar Q}{M} (1)

A continuación, analizamos cada caso:

Metano

Q = \frac{802\,\frac{kJ}{mol} }{16\,\frac{g}{mol} }

Q = 50.125\,\frac{kJ}{g}

1 gramo de metano aporta 50.125 kilojoules.

Octano

Q = \frac{5500\,\frac{kJ}{mol} }{114\,\frac{g}{mol} }

Q = 48.246\,\frac{kJ}{mol}

1 gramo de metano aporta 48.246 kilojoules.

3 0
2 years ago
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