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3 years ago

Rewrite and Balance the Equation N₂ + H₂ -> NH3

Chemistry

2 answers:

snow_lady [41]3 years ago

5 0

Answer:

you said brainliest

Explanation:

Mars2501 [29]3 years ago

4 0

Answer:N

+ 3

-----> 2N

Explanation:

-----> N

Let us balance this equation by counting the number of atoms on both sides of the arrow.

-----> N

N=2 , H=2 N=1, H=3

To balance the number of N atom on Right Hand Side (RHS) , I will add one molecule of N

on RHS

-----> 2N

N=2 , H=2 N=2 , H= 6

To balance the number of H atoms on Left Hand Side (LHS) , I will add two molecules of

on LHS

+ 3

-----> 2N

N=2 , H=6 N=2 , H= 6

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Which acid is the best choice to create a buffer with ph= 3.19?

Crazy boy [7]

<span>The best choice is hypochlorous acid nitrous acid (HNO2) because it has the nearest value of pK to the desired pH.
pKa of </span>nitrous acid<span> is 3.34
If we know pKa and pH values, we can calculate the required ratio of conjugate base (NO2⁻) to acid (HNO2) from the following equation:
pH=pKa + log(conc. of base)/( conc. of acid)
</span><span>3.19=3.34 + log c(NO2⁻)/c(HNO2)
</span><span>3.19 - 3.34 = log c(NO2⁻)/c(HNO2)
-0.15 = log c(NO2⁻)/c(HNO2)
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4 years ago

25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is

vlabodo [156]

Answer:

a) 119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

$\bigtriangleup T_f=K_fm, \ \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\$

#We use the molality above to calculate the molar mass:

$m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol$

Hence, the molar mass of the compound is 119 g/mol

5 0

4 years ago

The enthalpy of combustion of naphthalene (MW = 128.17 g/mol) is -5139.6 kJ/mol. How much energy is produced by burning 0.8210 g

bazaltina [42]

The energy produced by burning : -32.92 kJ

<h3>Further explanation</h3>

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The enthalpy and heat(energy) can be formulated :

$\tt \Delta H=\dfrac{Q}{n}\rightarrow n=mol$

The enthalpy of combustion of naphthalene (MW = 128.17 g/mol) is -5139.6 kJ/mol.

The energy released for 0.8210 g of naphthalene :

$\tt Q=\Delta H\times n\\\\Q=-5139.6~kJ/mol\times \dfrac{0.8210~g}{128.17~g/mol}~\\\\Q=-32.92~kJ$

3 0

3 years ago

The ionization constant (Ka) of HF is 6.7 X 10 ^-4. Which of the following is true in a 0.1 M solution of this acid?

spin [16.1K]

The ionization equation is:

HF ⇄ H(+) + F(-)

The ionization constant is Ka = [H(+)] * [H(-)] / [HF]

=> [H(+)] * [F(-)] = Ka * [HF]

Given that Ka < 1

[H(+)] * [F(-)] < [HF]

Which is [HF] > [H(+)] * [F(-)] the option a. fo the list of choices.

8 0

3 years ago

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Yakvenalex [24]

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3 years ago