Given 2.0 moles of nitrogen and plenty of hydrogen, how many moles of NH3 are formed?

Sydney and Valparaiso are the same distance from the equator and both are near the ocean. Is Valparaiso warmer, colder, or the s

PilotLPTM [1.2K]

Answer: It's colder.

Explanation: Well for starters Valparaiso is further away from the equator, and Australia is already really hot. But say that Valparaiso is further from the equator should be good enough.

5 0

3 years ago

Read 2 more answers

For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the v

Ierofanga [76]

Answer:

$\Delta G^o=-5.4032 kJ$

The temperature for $\Delta G^o=0[/tex is [tex]T=328.6 K$

Explanation:

The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:

$\Delta G^o=\Delta H^o + T*\Delta S^o$

Where:

$\Delta G^o$ is Gibbs's energy in kJ

$\Delta H^o$ is the enthalpy in kJ

$\Delta S^o$ is the entropy in kJ/K

$T$ is the temperature in K

Solving:

$\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K$

$\Delta G^o=-5.4032 kJ$

For $\Delta G^o=0$ :

$0=\Delta H^o - T*\Delta S^o$

$\Delta H^o= T*\Delta S^o$

$T=\frac{\Delta H^o}{\Delta S^o}$

$T=\frac{-58.03 kJ}{-0.1766 kJ/K}$

$T=328.6 K$

3 0

3 years ago

Read 2 more answers

CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?

Kipish [7]

<h3>Answer:</h3>

64 g O₂

<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>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 36 g H₂O

[Solve] x g O₂

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

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
Divide/Multiply [Cancel Units]: $\displaystyle 63.929 \ g \ O_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

63.929 g O₂ ≈ 64 g O₂

8 0

3 years ago

Read 2 more answers

a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill

baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

Also,

Moles = mass (m) / Molar mass (M)

Density (d) = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRT$

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>

5 0

3 years ago

The students in the picture below are using a globe and a lamp to model the Sun and the Earth. If the model Earth acts the same

timofeeve [1]

Answer:

The only one that makes sense IF the model behaves as the Earth is D.

Explanation:

7 0

3 years ago