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faltersainse [42]

2 years ago

12

Using the rxn below determine the following: What is the limiting reagent, theoretical yield of nitic oxide, and how many grams

of excess reagent remain when producing nitric oxide when using 4.65g of ammonia to react with 4.25g of oxygen?

1 answer:

WINSTONCH [101]2 years ago

4 0

5 mols of O_2 need 4 mol NH_3

As mols of ammonia are less Ammonia is limiting reagent

Now

Moles of Ammonia

Given mass/Molar mass
4.65/17
0.3mol

Moles of NO=0.3mol

Mass of NO

0.3(30)
9g

So

Moles of Oxygen

4.25/32
0.13mol

Remaining moles

0.3-0.13=0.17

Mass

0.17(32)
5.44g

Yield

5.44-4.25
0.79g

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1) A light bulb takes in 30 of energy per second. It transfers 3j as use

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Answer:

$\boxed{\text{10 \%}}$

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$\begin{array}{rcl}\text{Efficiency} & = & \dfrac{\text{useful energy out}}{\text{energy in}} \times 100 \,\% \\\\\eta & = & \dfrac{w_{\text{out}}}{w_{\text{in}} } \times 100 \,\%\\\end{array}$

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3 years ago

What is the change in internal energy (ΔΕ) of a system when 5 kJ of work is done on the system while it releases 13 kJ of energy

meriva

The change in internal energy (ΔΕ) of a system : -8 kJ

<h3>Further explanation </h3>

The laws of thermodynamics 1 state that: energy can be changed but cannot be destroyed or created

The equation is:

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The sign rules for heat and work are set as follows:

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• The system releases heat, Q -

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• the system accepts work, W +

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3 years ago

A substance with a mass of 50.0 g has a temperature change of 40.0

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2 years ago

Read 2 more answers

Write a balanced equation for the complete oxidation reaction that occurs when ethane burns in air

damaskus [11]

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General formula for alkanes :

$\tt \large{\bold{C_nH_{2n+2}}$

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$\large {\box {\bold{C_nH _ (_2_n _ + _ 2_) + \dfrac {3n + 1} {2} O_2 \Rightarrow nCO_2 + (n + 1) H_2O}}}$

So that the burning of ethane with air (oxygen):

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or we can use mathematical equations to solve equilibrium chemical equations by giving the coefficients for each compound involved in the reaction

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C : left 2, right b ⇒ b=2

H: left 6, right 2c⇒ 2c=6⇒ c= 3

O : left 2a, right 2b+c⇒ 2a=2b+c⇒2a=2.2+3⇒2a=7⇒a=7/2

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