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3 years ago

Write the word and balanced chemical equations for the reaction between:

Chemistry

1 answer:

nadezda [96]3 years ago

3 0

Answer:

3 HNO₃ + Fe(OH)₃ → H₂O + Fe(NO₃)₃

Explanation:

An acid reacts with a base producing water and a salt. Having this in mind the reaction of nitric acid (HNO₃) and Iron (III) hydroxide (Fe(OH)₃) is:

HNO₃ + Fe(OH)₃ → H₂O + Fe(NO₃)₃

The H⁺ of the acid reacts with the OH⁻ to produce H₂O. The other ions (Fe³⁺ and NO₃⁻) produce the salt

There are 3 nitrates in products. To balance the nitrates:

And this is the balanced reaction

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What is the mass percent of nitrogen in: 1. N2O 2. NO 3. NO2 4. N2O2

Temka [501]

We’re going to use the mass percent formula shown below:

For the percent by mass N, we’re going to rewrite the equation as:

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Problem Details
Calculate the mass percent composition of nitrogen in each nitrogen-containing compound:

c. NO2

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2 years ago

A rigid vessel contains 3.98 kg of refrigerant-134a at 700 kPa and 60°C. Determine the volume of the vessel and the total intern

Arada [10]

Answer: The volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

Explanation:

To calculate the volume of water, we use the equation given by ideal gas, which is:

$PV=nRT$

or,

$PV=\frac{m}{M}RT$

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of container = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L$

Converting this value into $m^3$ , we use the conversion factor:

$1m^3=1000L$

So, $\Rightarrow (\frac{1m^3}{1000L})\times 154.21L$

$\Rightarrow 0.1542m^3$

To calculate the internal energy, we use the equation:

$U=\frac{3}{2}nRT$

or,

$U=\frac{3}{2}\frac{m}{M}RT$

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.314J/K.mol$

T = temperature = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

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Explanation:

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