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WINSTONCH [101]
3 years ago
11

Hydrogen and oxygen react chemically to form water how much water would form if 14.8grams of hydrogen reacted with 34.8 grams of

oxygen (H2+O2->H2O)
Chemistry
1 answer:
pishuonlain [190]3 years ago
4 0

Answer:

There will be formed 39.1935 grams H2O formed

Explanation:

<u>Step 1:</u> The balanced equation

2H2 + 02 → 2H20

<u>Step 2</u>: Given data

mass of hydrogen = 14.8 grams

Molar mass of hydrogen = 2.02 g/mole

mass of oxygen = 34.8 grams

Molar mass of oxygen = 32 g/mole

<u>Step 3: </u>Calculate moles

moles = mass / Molar mass

moles of hydrogen = 14.8g/ 2.02 g/mole = 7.33 moles

moles of oxygen = 34.8g / 32g/mole = 1.0875 moles

For 2 moles hydrogen consumed, we need 1 mole of oxygen.

This means oxygen is the limiting reagens and will be consumed completely. Hydrogen is the reactant in excess, there will remain 5.155 moles of hydrogen

<u>Step 4:</u> Calculate moles of H2O

We see that for 2 moles of H2 consumed, there is needed 1 mole of O2, to produce 2 moles of H2O.

For 1.0875 moles of oxygen consumed, there will be produced 2.175 moles of H2O

<u>Step 5:</u> Calculate mass of water

Mass of H2O = moles of H2O * Molar mass of H2O

Mass of H2O = 2.175 moles * 18.02 g/moles 39.1935 grams

There will be formed 39.1935 grams H2O formed

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Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163
statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

5 0
3 years ago
What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
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devlian [24]
The state of matter with the slowest moving atoms is solid!
6 0
3 years ago
Why do materials interact to light differently?
liubo4ka [24]

Answer:

Refracted (refraction) – the bending of transmitted light as it travels across the boundary of one material into another material in which it's speed is different. Unlike diffraction, this change in direction of light occurs because light is changing its speed in the different substances in it traveling in.

Explanation:

Hope it helps

6 0
3 years ago
Balance the following H2 SO4 + NaOH ==== NaSO + H2O​
katrin [286]

Answer:

H2SO4 + 2NaOH --->   Na2SO4 + 2H2O

Explanation:

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H2SO4 + 2NaOH --->   Na2SO4 + H2O

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H2SO4 + 2NaOH --->   Na2SO4 + 2H2O

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