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3 years ago

What is the balance of NH3+HCL=NH4CL+H2O

Chemistry

1 answer:

Karolina [17]3 years ago

3 0

Answer:

Hydrogen chloride react with ammonium hydroxide

HCl + NH3•H2O → NH4Cl + H2O

Explanation:

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Which of these best describes the state of the substance represented by State Z?

romanna [79]

A) it is a solid because strong attractive forces prevent particles from moving.

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3 years ago

In the reversible reaction: 2NO2 (g) ⇌ N2O4 (g) the formation of dinitrogen tetroxide releases heat and the formation of nitroge

xenn [34]

Answer: Option (B) is the correct answer.

Explanation:

According to Le Chatelier's principle, any disturbance causes in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

For example, $2NO_{2}(g) \rightleftharpoons N_{2}O_{4}(g)$

When we increase the temperature then the reaction will shift in a direction where there will be decrease in temperature.

This, means that the reaction will shift in the backward direction.

Thus, we can conclude that if the reaction is at equilibrium and the temperature increases, the equilibrium will shift so that there is more nitrogen dioxide.

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3 years ago

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2.5 moles of gas occupies 50 L, how many moles of the gas would occupy 100L?

goldenfox [79]

Answer:

Explanation:

2.5M=50L

X =100L

2.5MX100L

________ =250M/L

5OL 50L

=5 Moles

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3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

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3 years ago

What is the mass of one mole of nitrogen atoms?

OLEGan [10]

Answer:

28.0

Explanation:

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3 years ago

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