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IRINA_888 [86]
4 years ago
8

How many molecules of ammonia are formed in the following equation: N2+3H2→2NH3

Chemistry
1 answer:
Katena32 [7]4 years ago
7 0

Answer:

  • <u>two molecules of ammonia are formed by the reaction of one nitrogen and three hydrogen molecules.</u>

Explanation:

The  balanced chemical equation provides information on:

  • <u>Reactants</u>: those are the compounds that appear of the left side of the equation, each with its chemical formula.

  • <u>Products</u>: those are the compounds that appear on the right side of the equation, again, each with its chemical formula.

  • <u>Ratio</u>: the coefficients of each compound (the number to the left of the chemical formula) represent the ratio of the number of molecules that react and are formed.

In the given equation you have:

  • Equation: N₂ + 3H₂ → 2NH₃

  • The coefficients are 1 for nitrogen, 3 for hydrogen, and 2 for ammonia. Hence, 2 molecules of ammonia are formed by the reaction of 1 molecule of nitrogen and 3 molecules of hydrogen.
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Predict the products for the following equation.
coldgirl [10]

Answer: AgCl + Fe

Explanation:

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A general single displacement reaction can be represented as :

XY+Z\rightarrow XZ+Y

The blanced chemical reaction will be :

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Mercury is a liquid meteal but it is not hard? how can we say it is a metal
NNADVOKAT [17]
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3 years ago
The smell of fresh cut pine is due in part to the cyclic alkene called pinene. Given the following data of pinene: Vapor pressur
Sveta_85 [38]

Answer: Heat of vaporization is 41094 Joules

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1= initial pressure at  429 K = 760 torr

P_2 = final pressure at 415 K  = 515 torr

= enthalpy of vaporisation = ?

R = gas constant = 8.314 J/mole.K

T_1= initial temperature =  429 K

T_2 = final temperature = 515 K

Now put all the given values in this formula, we get

\log (\frac{515}{760}=\frac{\Delta H}{2.303\times 8.314J/mole.K}[\frac{1}{429K}-\frac{1}{415K}]

\Delta H=41094J

Thus the heat of vaporization is 41094 Joules

4 0
3 years ago
Problem Page Suppose 25.1g of barium nitrate is dissolved in 200.mL of a 0.50 M aqueous solution of ammonium sulfate. Calculate
melomori [17]

Answer:

1.26 M

Explanation:

The ion nitrate is NO₃⁻ and the Barium is from group 2 so it forms the ion Ba²⁺, so the barium nitrate has the formula: Ba(NO₃)₂. The molar masses are: Ba: 137 g/mol, N = 14 g/mol, O = 16 g/mol, so the molar mass of barium nitrate is:

137 + 2x(14 + 3x16) = 199 g/mol

The number of moles is the mass divided by the molar mass, so:

n = 25.1/199 = 0.126 mol of Ba(NO₃)₂

In 1 mol of the salt, there are 2 moles of NO₃⁻, so the number of moles of nitrate is 0.252 mol. Nitrates formed with ammonium (that can react when the solid dissolves) and with elements from group 1 and 2 are completely soluble in water. So, the moles of nitrate will remain 0.252 mol.

The molarity is the number of moles divided by the volume (0.2 L):

[NO₃⁻]= 0.252/0.2 = 1.26 M

3 0
3 years ago
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