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3 years ago

At STP, which sample contains the same number of molecules as 3.0 liters of hydrogen gas?

Chemistry

1 answer:

uranmaximum [27]3 years ago

5 0

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Organic chemistry please help

lord [1]

The answer to this is C. III & IV

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4 years ago

The water gained 900 J of energy, what was the energy change of the metal?

Gemiola [76]

Answer:

it lost 900J of energy I will assume and that would mean the change in energy is -900J

Explanation:

4 0

3 years ago

How many grams of RbOH are present in 35.0 mL of a 5.50 M RbOH solution?

Ganezh [65]

I think 5.50 M x 35.0 mL x molar mass of RbOH = mass (g)

8 0

4 years ago

Value of Δ H ∘ rxn for the equation NH 3 ( g ) + 2 O 2 ( g ) ⟶ HNO3 ( g ) + H 2 O ( g )

koban [17]

Answer: - 894.6 kJ/mol.

Explanation:

Hess law is states that the changes in enthalpies in a chemical reactions is independent of the pathway between the initial and final states.

∆H is the change in the sum of the internal energy of a system.

We are to find the Value of ΔH°(rxn) for the equation:

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ----------------------------------(**).

From the series of equations given;

==> 4NH3 (g) + 5O2 (g) -------->4 NO(g) + 6H2O (l). ∆H = -1166.0 kJ/mol.--------------------------------------(1).

===> 2NO(g) + O2 (g) ------> 2NO2 (g). ∆H = -116.2 kJ/mol.---------------(2).

===> 3NO2 (g) + H2O (l) ---------> 2HNO3 (aq) + NO (g). ∆H = -137.3 kJ/mol.-------------------------------------(3).

The first thing to do is to multiply equation (2) by 3. Also, multiply equation (3) by 2. This will give us equation (4) and (5) respectively.

6NO + 3 O2 ----------------> 6NO2. ∆H= 3 × (-116.2 kJ/mol) = -348.6 kJ/mol. ------------------------------------(4).

6NO2 + 2 H2O ----------------> 4HNO3 + 2 NO. ∆H= 2 × (-137.3kj/mol) = -274.6 kJ/mol ---------------------------(5).

Next, add equations (4) and (5) to give;

4NO +3O2 +2H2O -------------> 4HNO3. ∆H = -623.2 kJ/mol. -----(6).

Add this equation to the equation (1) from above, we have;

4NH3 + 8O2 --------------> 4HNO3 + 4H2O. ∆H= -1789.2 kJ/mol. --------(7).

Then, divide the equation (7) above by 2 to give us back the equation (**).

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ∆H= -894.6 kJ/mol.

Δ H^∘ (rxn)= - 894.6 kJ/mol.

5 0

3 years ago

What mass (g) of magnesium nitride (Mg3N2) can be made from the reaction of 1.22 g of magnesium with excess nitrogen? __Mg + __N

Citrus2011 [14]

<h3>Answer:</h3>

1.69 g Mg₃N₂

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis
Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

<u>Step 2: Identify Conversions</u>

[RxN] 3 mol Mg → Mg₃N₂

[PT] Molar Mass of Mg - 24.31 g/mol

[PT] Molar Mass of N - 14.01 g/mol

Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 1.22 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{1 \ mol \ Mg_3N_2}{3 \ mol \ Mg})(\frac{100.95 \ g \ Mg_3N_2}{1 \ mol\ Mg_3N_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 1.68873 \ g \ Mg_3N_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

8 0

3 years ago