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3 years ago

what is the reason for the difference between the values of the dry adiabatic lapse rate and the moist adiabatic lapse rate?

1 answer:

6 0

Latent heat released during condensation offsets some of the cooling of a rising air parcel, causing the moist adiabatic lapse rate to have a smaller value.

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Evaluate this statement substances are made of two or more types of elements

lapo4ka [179]

The substance is a compound

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4 years ago

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What is the coronavirus? Where was it first found in?

Lena [83]

Answer:

The corona virus is a type of drug jk it is a deadly fast spreading virus that was first discovered in Wuhan China I believe

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3 years ago

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Which of the following can be used to neutralize a sodium hydroxide (NaOH) solution? NaCl NH3 HCl Ca(OH)2

Andru [333]

NaOH is a base and hence it can only be neutralized by an acid.

So it is by HCl, which is an acid.

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3 years ago

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When adding the measurements 42.1014 g + 190.5 g, the answer has ___ significant figures.

andrezito [222]

Answer: The answer has 7 significant figures

Explanation:

The addition of 190.5 and 42.1014 will give 232.6014. Counting the digits will give 7 significant figures.

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4 years ago

g The following reaction is the first step in the production of nitric acid from ammonia. 4NH3(g) 5O2(g) → 4NO(g) 6H2O(g) Calcul

STatiana [176]

<u>Answer:</u> The enthalpy of the reaction is coming out to be -902 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(4\times \Delta H_f_{(NO(g))})+(6\times \Delta H_f_{(H_2O(g))})]-[(4\times \Delta H_f_{(NH_3(g))})+(5\times \Delta H_f_{(O_2)})]$

We are given:

$\Delta H_f_{(NO(g))}=91.3kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-45.9kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(4\times (91.3))+(6\times (-241.8))]-[(4\times (-45.9))+(5\times (0))]\\\\\Delta H_{rxn}=-902kJ$

Hence, the enthalpy of the reaction is coming out to be -902 kJ.

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4 years ago