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kherson [118]
3 years ago
13

Mass divided by volume *

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
5 0
D — density

hope this helps !
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Why was the term inert gases dropped? Why was the term inert gases dropped? The term "inert" was dropped because hydrogen was ad
xxMikexx [17]

Answer:

B) The term "inert" was dropped because it no longer described all the group 8A elements.

Explanation:

Inert elements in chemistry simply refers to elements that are chemically inactive and are not expected to form any compounds. this is the general belief for the group 8 elements as they all have complete duplet/octet configurations (and ideally, they ought to be very stable with no tendency to form compounds by participating in the loss and gain of electrons). However the discovery of compounds like xenon tetrafluoride (XeF4) proved this to be wrong.

Again, the reason the term - inert gses was droppedis beacause this term is not strictly accurate because several of them do take part in chemical reactions.

After dropping the term - Inert gases, they are now referred to as noble gases.

8 0
3 years ago
Which of these best describes a scientific model?
goblinko [34]
The answer a way of explaining a complex concept.
3 0
3 years ago
¿Todos los solutos se disuelven en cualquier solvente?
Katen [24]

Answer:

No

Explanation: No. Todos los solutos tienen una solubilidad diferente, lo que significa que todos tienen una cantidad única de gramos que se disolverían en el mismo solvente. una solución que ha disuelto todo el soluto que puede contener a una temperatura determinada.

6 0
3 years ago
The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia
Savatey [412]

Answer:

\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0
3 years ago
What mass of barium chloride, BaCL2, is required to completely precipitate the barium sulfate from 35.0 ml of 0.200 m h2so4?
faltersainse [42]
The balanced equation would be (1)BaCl2 + (1)H2SO4 --> (1)BaSO4 + (2)HCl2
Then you should know that the coefficients stand for moles.
The thing is I'm not sure if H2SO4 is 35 ml or .200 m. 
Also, is this topic stoichiometry?
5 0
3 years ago
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