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3 years ago

Balance the following chemical equation:N2H4+O2 =NO2 +H2O

Chemistry

2 answers:

Deffense [45]3 years ago

4 0

The answer for the following question is mentioned below.

$N_{2}H_{4}$ +3 $O_{2}$ →2 $NO_{2}$ +2 $H_{2} O$

the equation is balanced.

Explanation:

The steps required to balance a chemical equation is:

Firstly,write the formula of the compound/substance.

Secondly,count the number of atoms in each element on each side of the equation

Thirdly,Adjust the co-efficient.Balance the atoms or the ions present in the greatest number.

You may do this by finding the lowest common multiple of two.

Continue systematically to balance the atoms or ions on the rest of the side (i.e.)on the other side.

Lastly,check the number of atoms each elements is same on both the sides

From the equation:

Before balancing the equation:

$N_{2} H_{4}$ + $O_{2}$ → $NO_{2}$ + $H_{2} O$

After balancing the equation:

$N_{2}H_{4}$ +3 $O_{2}$ →2 $NO_{2}$ +2 $H_{2} O$

Therefore the given equation is balanced.

hram777 [196]3 years ago

4 0

Hey there!

N₂H₄ + O₂ → NO₂ + H₂O

Balance N.

2 on the left, 1 on the right. Add a coefficient of 2 in front of NO₂.

N₂H₄ + O₂ → 2NO₂ + H₂O

Balance H.

4 on the right, 2 on the right. Add a coefficient of 2 in front of H₂O.

N₂H₄ + O₂ → 2NO₂ + 2H₂O

Balance O.

2 on the left, 6 on the right. Add a coefficient of 3 in front of O₂.

N₂H₄ + 3O₂ → 2NO₂ + 2H₂O

Our final equation: N₂H₄ + 3O₂ → 2NO₂ + 2H₂O

Hope this helps!

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3 years ago

Without consulting Appendix B, arrange each group in order of increasing standard molar entropy (S°). Explain.(c) SF₆(g), SF₄(g)

Andre45 [30]

The increasing order of standard molar entropy (S°) is as follow:

SF₄(g) < SF₆(g) < S₂F₁₀(g)

<h3>What is Entropy? </h3>

Entropy is defined as the randomness of the particle. It depends on temperature and pressure or number of particle per unit volume.

It is directly proportional to the temperature and pressure of the gas.

<h3>What is Standard Molar Entropy? </h3>

The standard molar entropy is defined as the entropy content of the one mole of pure substance at the standard state of temperature and pressure of interest.

The standard molar entropy is also defined as the total amount of entropy which 1 mole of the substance acquire, as it is brought from 0K to standard conditions of temperature and pressure.

The standard molar entropy depends on the molas mass of atom, molecules or compound.

SF₄(g) has lower standard molar entropy. Due to less complexity of this molecules.

While, complexity increases from SF₆(g) to S₂F₁₀(g). Therefore, the standard molar entropy of S₂F₁₀(g) is greater than SF₆(g).

Thus, we concluded that the increasing order of standard molar entropy (S°) is as follow:

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2 years ago

All of the following are true about the Earth’s inner core EXCEPT

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4 years ago

What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m

djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL (1cm³ = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³ mol / 0.4327 L = 0.0075 M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x 432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

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3 years ago

Balance the following chemical equation:__N2H4+__O2 =__NO2 +__H2O​

Balance the following chemical equation:N2H4+O2 =NO2 +H2O