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elixir [45]
3 years ago
6

For the following reaction, 5.21 grams of nitrogen gas are allowed to react with 5.91 grams of hydrogen gas.

Chemistry
1 answer:
xz_007 [3.2K]3 years ago
7 0

Answer:

Maximum amount of ammonia that can be formed is 6.32 g

Explanation:

The balanced equation is:

N2 + 3H2\rightarrow 2NH3

Moles\ N2 = \frac{5.21g}{28g/mol} =0.186\\\\Moles H2 = \frac{5.21g}{2g/mol} =2.61

N2 is the limiting reactant

Based on the reaction stoichiometry:

1 mole N2 produced 2 moles NH3

therefore, 0.186 mol N2 will yield: 2(0.186) = 0.372 moles NH3

Molar mass NH3 = 17 g/mol

Maximum\ Mass\ NH3 = 0.372 mols*17g/mol=6.32g

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c ammonium hydroxide nh4oh has higher ph value and its a base

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3 years ago
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A 0.08541 g sample of gas occupies 10.0-ml at 288.5 k and 1.10 atm. upon further analysis, the compound is found to be 13.068% c
topjm [15]
<span>C2Br2 First, we need to determine how many moles of the gas we have. For that, we'll use the Ideal Gas Law which is PV = nRT where P = pressure (1.10 atm = 111458 Pa) V = volume (10.0 ml = 0.0000100 m^3) n = number of moles R = Ideal gas constant (8.3144598 (m^3 Pa)/(K mol) ) T = Absolute temperature Solving for n, we get PV/(RT) = n Now substituting our known values into the formula. (111458 Pa * 0.0000100 m^3) / (288.5 K * 8.3144598 (m^3 Pa)/(K mol)) = (1.11458/2398.721652) mol = 0.000464656 mol Now let's calculate the empirical formula for this compound. Atomic weight carbon = 12.0107 Atomic weight bromine = 79.904 Relative moles carbon = 13.068 / 12.0107 = 1.08802984 Relative moles bromine = 86.932 / 79.904 = 1.087955547 So the relative number of atoms of the two elements is 1.08802984 : 1.087955547 After dividing all numbers by the smallest, the ratio becomes 1.000068287 : 1 Which is close enough to 1:1 for me to consider the empirical formula to be CBr Now calculate the molar mass of CBr 12.0107 + 79.904 = 91.9147 Finally, let's determine if the compound is actually CBr, or something like C2Br2, or some other multiple. Using the molar mass of CBr, multiply by the number of moles and see if the result matches the mass of the gas. So 91.9147 g/mol * 0.000464656 mol = 0.042708701 g 0.0427087 g is a lot smaller than 0.08541 g. So the compound isn't exactly CBr. Let's divide them to see what the factor is. 0.08541 / 0.0427087 = 1.99982673 1.99982673 is close enough to 2 to within the number of significant digits we have for me to claim that the formula for the unknown gas isn't CBr, but instead is C2Br2.</span>
3 0
3 years ago
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During a reaction, the enthalpy of formation of an intermediate is 34 kJ/mol.
galina1969 [7]

Since the enthalpy can be calculated from the heat of formation, the enthalpy is 136 kJ/mol.

<h3>What is enthalpy?</h3>

The enthalpy of a reaction is the heat that is lost or gained in that reaction. We know that the enthalpy can be calculated from the heat of  formation.

Thus, we can obtain the enthalpy of the reaction as 4 * 34 kJ/mol = 136 kJ/mol.

Learn more about enthalpy:brainly.com/question/13996238

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5 0
2 years ago
CO2 is an element. True False
Reil [10]

Answer:False

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1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely n
Ede4ka [16]

Answer:

m_{H2SO4 = 61.25 g

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2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

   2        :     1           :      1         :    2

 1.25                                                       (moles)

⇒  n_{H2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ m_{H2SO4} = 0.625 × 98 = 61.25 g

    n_{Na2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒m_{Na2SO4} = 0.625 × 142 = 88.75 g

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