It take more energy to break the bonds of the reactants and less energy is given off when the product bonds are formed.
<h3>What is Energy?</h3>
Energy is defined as the ability to do work. Work is done in the breaking or formation of bonds.
The standard Enthalpy (ΔH) of water which was formed in the given reaction is negative.
ΔH= Δproduct - Δreactant
This means that the energy to break the bonds of the reactants is more.
Read more about Enthalpy here brainly.com/question/14291557
Answer:
15.0 L
Explanation:
To find the volume, you need to use the Ideal Gas Law:
PV = nRT
In this equation,
-----> P = pressure (mmHg)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)
-----> T = temperature (K)
To calculate the volume, you need to (1) convert grams C₄H₁₀ to moles (via the molar mass), then (2) convert the temperature from Celsius to Kelvin, and then (3) calculate the volume (via the Ideal Gas Law).
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
32 grams C₄H₁₀ 1 moles
------------------------- x ----------------------- = 0.551 moles C₄H₁₀
58.124 grams
P = 728 mmHg R = 62.36 L*mmHg/mol*K
V = ? L T = 45.0 °C + 273.15 = 318.15 K
n = 0.551 moles
PV = nRT
(728 mmHg)V = (0.551 moles)(62.36 L*mmHg/mol*K)(318.15 K)
(728 mmHg)V = 10922.7632
V = 15.0 L
Answer:
0.0917 mol Co(CrO₄)₃
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
37.3 g Co(CrO₄)₃
<u>Step 2: Identify Conversions</u>
Molar Mass of Co - 58.93 g/mol
Molar Mass of Cr - 52.00 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Co(CrO₄)₃ - 58.93 + 3(52.00) + 12(16.00) = 406.93 g/mol
<u>Step 3: Convert</u>
<u />
= 0.091662 mol Co(CrO₄)₃
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.091662 mol Co(CrO₄)₃ ≈ 0.0917 mol Co(CrO₄)₃
The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>
Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
Mol ratio of reactants(to find limiting reatants) :
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
Answer:
Option B. 176g/mol
Explanation:
We'll begin by writing the chemical formula for hexasodium difluoride. This is given below:
Hexasodium means 6 sodium atom
Difluoride means 2 fluorine atom.
Therefore, the formula for hexasodium difluoride is Na6F2.
The relative formula mass of a compound is obtained by simply adding the atomic masses of the elements present in the compound.
Thus, the relative formula mass of hexasodium difluoride, Na6F2 can be obtained as follow:
Molar mass of Na = 23g/mol
Molar mass of F = 19g/mol
Relative formula mass Na6F2 = (23x6) + (19x2)
= 138 + 38
= 176g/mol
Therefore, the relative formula mass of hexasodium difluoride, Na6F2 is 176g/mol
