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2 years ago

Assume the following substances are soluble in water.

Chemistry

1 answer:

solmaris [256]2 years ago

8 0

Ammonium carbonate will form 3 moles of ions.

Methyl alcohol will form 0 moles of ions.

Methane will form 0 moles of ions.

Aluminum sulfite will form 3 moles of ions.

Hydrobromic acid will form 2 moles of ions.

Explanation:

One mole of ammonium carbonate will form 3 moles of ions when dissolved in water.

(NH₄)₂CO₃ (s) + H₂O (l) → 2 NH₄⁺ (aq) + CO₃²⁻ (aq) + H₂O (l)

One mole of methyl alcohol will form 0 moles of ions when dissolved in water.

(actually it form ions because of its acidic behavior but they are at the order of 10⁻⁷ moles, however in the framework of this question we may say that there are none)

One mole of methane will form 0 moles of ions when dissolved in water.

Methane does not react with water (in normal conditions) so will not form ions.

One mole of aluminum sulfite will form 3 moles of ions when dissolved in water.

Al₂SO₃ (s) + H₂O (l) → 2 Al₃⁺ (aq) + SO₃²⁻ (aq) + H₂O (l)

One mole of hydrobromic acid will form 2 moles of ions when dissolved in water.

HBr (l) + H₂O (l) → Br⁻ (aq) + H₃O⁺ (aq)

Learn more about:

solvation of ions

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Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t

mars1129 [50]

Answer:

Explanation:

N₂ + 3H₂ = 2 NH₃

1 vol 2 vol

786 liters 1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

$\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }$

$V_2 =303.44$ liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

=.23 kg / sec .

Rate of production of ammonia = .23 kg /s .

5 0

2 years ago

How many grams of NO are required to produce 145 g of N2 in the following reaction?

V125BC [204]

Answer:

b. 186 g

Explanation:

Step 1: Write the balanced equation.

4 NH₃(g) + 6 NO(g) → 5 N₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 145 g of N₂

The molar mass of nitrogen is 28.01 g/mol.

$145g \times \frac{1mol}{28.01 g} =5.18 mol$

Step 3: Calculate the moles of NO required to produce 5.18 moles of N₂

The molar ratio of NO to N₂ is 6:5.

$5.18molN_2 \times \frac{6molNO}{5molN_2} = 6.22molNO$

Step 4: Calculate the mass corresponding to 6.22 moles of NO

The molar mass of NO is 30.01 g/mol.

$6.22mol \times \frac{30.01g}{mol} =186 g$

4 0

1 year ago

Industrial production of nitric acid, which is used in many products including fertilizers and explosives, approaches 10 billion

mylen [45]

Answer: $9.361\times 10^{4} kJ$

Explanation:

The balanced chemical equation :

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$ $\Delta H^0_{rxn}=-902.0kJ$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.056\times 10^3g}{17g/mol}=415.1moles$

According to stoichiometry:

4 moles of $NH_3$ produces = 902.0 kJ of energy

415.1 moles of $NH_3$ produces = $\frac{902.0}{4}\times 415.1=9.361\times 10^{4} kJ$ of energy

Thus the change in enthalpy is $9.361\times 10^{4} kJ$

5 0

2 years ago

What is the percent composition by mass of nitrogen in the compound N2H4 (gram-formula mass = 32 g/mol)?

Alex

Answer:

$\large \boxed{93\, \% }$

Explanation:

1. Calculate the molar mass of N₂H₄

2N = 2 × 14 = 28

4H = 2 × 1 = <u> 4</u>

Tot. = 32

2. Calculate the mass percent of N

$\text{\% of element} = \dfrac{\text{mass of element}}{\text{mass of compound}} \times 100 \, \% = \dfrac{\text{28}}{\text{32}} \times 100 \, \% =\mathbf{88 \, \%}\\\\\text{The percentage of N in N$_{2}$H$_{4}$ is $\large \boxed{\mathbf{88\, \% }}$}}$