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Angelina_Jolie [31]
3 years ago
7

When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas write the balanved equati

on
Chemistry
1 answer:
babunello [35]3 years ago
6 0

Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}.

Explanation:

The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.

Now, in terms of chemical formulae this reaction equation will be as follows.

MnO_{2} + HCl \rightarrow H_{2}O + MnCl_{2} + Cl_{2}

Here, number of atoms on reactant side are as follows.

  • Mn = 1
  • O = 2
  • H = 1
  • Cl = 1

Number of atoms on product side are as follows.

  • Mn = 1
  • O = 1
  • H = 2
  • Cl = 4

To balance this equation, multiply HCl by 4 on reactant side and multiply H_{2}O by 2 on product side. Therefore, the equation can be rewritten as follows.

MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}

Hence, number of atoms on reactant side are as follows.

  • Mn = 1
  • O = 2
  • H = 4
  • Cl = 4

Number of atoms on product side are as follows.

  • Mn = 1
  • O = 2
  • H = 4
  • Cl = 4

Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.

Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}.

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Which types of reactions would result in a change in temperature?
Law Incorporation [45]

Answer:

Chemical reaction involves the breaking of bonds in the reactants and formation of bonds in the products. ... If a reaction is exothermic, more energy is released when the bonds of the products are formed than it takes to break the bonds of the reactants. This is the reason for temperature change during a reaction.

Explanation:

Here are just a few everyday demonstrations that temperature changes the rate of chemical reaction: Cookies bake faster at higher temperatures. Bread dough rises more quickly in a warm place than in a cool one.

6 0
3 years ago
Rank the following fertilizers in decreasing order of mass percentage of nitrogen:
charle [14.2K]
<h3>Answer:</h3>

        NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

<h3>Soution:</h3>

In (NH₄)₂HPO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄  =  132.06 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age  =  21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age  =  21.18 %

In KNO₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of KNO₃  =  101.10 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of KNO₃ × 100

Mass %age  =  14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age  =  13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄  =  115.03 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age  =  14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age  =  12.17 %

In NH₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of NH₃  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₃ × 100

Mass %age  =  14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age  =  82.20 %

In NH₄NO₃:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of NH₄NO₃  =  80.04 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age  =  28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

Mass %age  =  34.98 %

5 0
3 years ago
While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e
sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                  1.8                4.7

<u>At eqllm:</u>           1.8-x             4.7-x

Evaluating the value of 'x'

\Rightarrow (1.8-x)=1.16\\\\x=0.64

The expression of K_p for above equation follows:

K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}

p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm

Putting values in above expression, we get:

K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                2.36             4.06               0.64

<u>At eqllm:</u>           2.36-x        4.06-x             0.64+x

Putting value in the equilibrium constant expression, we get:

0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

3 0
3 years ago
A certain hydrocarbon has a molecular formula of C9H16. Which of the following is not a structural possibility for this hydrocar
snow_tiger [21]

Answer: Option (d) is the correct answer.

Explanation:

It is given that molecular formula is C_{9}H_{16}. Now, we will calculate the degree of unsaturation as follows.

Degree of unsaturation = C_{n} - \frac{\text{monovalent}}{2} + \frac{\text{trivalent}}{2} + 1

                                  = 9 - \frac{16}{2} + 1

                                  = 9 - 8 + 1

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As the degree of unsaturation comes out to be 2. It means that this compound will contain one ring and one double bond.

Yes, this compound could be an alkyne as for alkyne D.B.E = 2.

But this compound cannot be a cycloalkane because for a cycloalkane D.B.E = 1 which is due to the ring only.

Thus, we can conclude that it is a cycloalkane is not a structural possibility for this hydrocarbon.

6 0
3 years ago
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First we have to understand what a half life is. A half life is a measure of time when the amount of a certain object is 50% of the original amount. Hence the answer for this is letter A. The initial amount is 100. fifty percent of 100 is 50 and that happens after 15 hours. Hence, 15 hours is the half life period. 
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3 years ago
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