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umka2103 [35]
3 years ago
7

How many moles are in 20 grams of Ar?​

Chemistry
1 answer:
pashok25 [27]3 years ago
7 0

Answer:

moles Ar in 20g = 0.500 mole Ar

Explanation:

moles = grams given / formula weight = 20g / 39.948g·mol⁻¹ = 0.500 mole Ar

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The measurement 1.00540 g contains<br> significant figures. *<br> 10<br> 3<br> 4<br> 5<br> O<br> 6
antiseptic1488 [7]
It contains 6 sig figs
4 0
3 years ago
You run an experiment in which two substances chemically react in a closed system. you run the same experiment in an open system
m_a_m_a [10]

If we run an experiment in which two substances chemically react in a closed system and then we run the same experiment in an open system, then the masses of the products in each experiment will be different because the open system allows the interchange of matter and energy with the media.

<h3>What is an open system?</h3>

An open system is an interrelated group pf parts that work together to interchange matter and energy with the surrounding environment, while a closed system does not generate an exchange of matter and energy.

Therefore, with this data, we can see that an open system interchange energy and matter with the environment, while closes systems do it.

Learn more about open systems here:

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6 0
1 year ago
For the reaction where Δn=−1Δn=−1 , what happens after in increase in volume? ????KQ&gt;K so the reaction shifts toward reactant
Alex787 [66]

Answer:

Explanation:

In general, an increase in pressure (decrease in volume) favors the net reaction  that decreases the total number of moles of gases, and a decrease in pressure (increase in volume) favors the net reaction that increases  the total number of moles of gases.

Δn= b - a

Δn=  moles of gaseous products - moles of gaseous reactants

Therefore, <u>after the increase in volume</u>:

  • If Δn= −1 ⇒ there are more moles of gaseous reactants than gaseous products. The equilibrium will be shifted towards the products, that is, from left to right, and K>Q.
  • If Δn= 0 ⇒ there is the same amount of gaseous moles, both in products and reactants. The system is at equilibrium and K=Q.
  • Δn= +1 ⇒ there are more moles of gaseous products than gaseous reactants. The equilibrium will be shifted towards the reactants, that is, from right to left, and K<Q.

8 0
3 years ago
I need help ASAP please help me now​
Neko [114]

Answer:

it has to be a no cap !! buh

3 0
3 years ago
Read 2 more answers
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
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