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hjlf
2 years ago
11

Classify the molecules based on whether the molecule has a standard enthalpy of formation, AH;, equal to 0. Assume all

Chemistry
1 answer:
Semmy [17]2 years ago
4 0

The standard enthalpy of formation of substance in its standard state is zero.

Enthalpy is defined as a thermodynamic quantity that describes the energy of a system. For substances in their standard state, the enthalpy of formation is zero.

The standard state of a substance is defined as the state in which it is found under standard conditions.  The following substances has their standard ethalpy of formation as zero or not zero;

Zero enthalpy of formation           Non zero enthalpy of formation

Cl2(g)                                                     I2(s)

Br(g)                                                       Br2(l)

I2(g)                                                       Br2(s)

Hg(l)                                                       Hg(s)

Learn more: brainly.com/question/13164006

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How many grams of zinc phosphate are formed when 48.1 mL of 2.18 M zinc nitrate reacts with excess potassium phosphate?
Roman55 [17]

Answer:

15.4 g of Zn₃(PO₄)₂ are produced

Explanation:

Given data:

Mass of zinc phosphate formed = ?

Volume of zinc nitrate = 48.1 mL (0.05 L)

Molarity of zinc nitrate = 2.18 M

Solution:

Chemical equation:

3Zn(NO₃)₂ + 2K₃PO₄  →   Zn₃(PO₄)₂ + 6KNO₃

Moles of  zinc nitrate:

Molarity = number of moles / volume in litter

Number of moles =  2.18 M × 0.05 L

Number of moles = 0.109 mol

Now we will compare the moles of zinc phosphate with zinc nitrate from balanced chemical equation:

                     Zn(NO₃)₂           :            Zn₃(PO₄)₂

                         3                    :                 1

                       0.109               :               1/3×0.109 = 0.04 mol

0.04 moles of Zn₃(PO₄)₂ are produced.

Mass of Zn₃(PO₄)₂:

Mass = number of moles × molar mass

Mass = 0.04 mol × 386.1 g/mol

Mass = 15.4 g

   

6 0
2 years ago
A sample of an unknown gas at STP has a density of 0.630 gram per liter. What is the gram molecular mass of this gas?
jeyben [28]
<span>STP means standard temperature and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the numer of moles and can be represented as mass of the gas, m, divided by the molar mass, c.  so we have,</span>  

PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
c = drT/P  

substitute the values into the equation,
c = [(0.63g/L)(0.08206 L-atm/mol-K)(273K)]/(1atm)
<u>c = 14.11 g/mol</u>
3 0
3 years ago
Joy mixes one cup of sugar and one cup of lemon juice into three cups of water. The solvent in this recipe is the?
kobusy [5.1K]

Answer: Water is the solvent in this recipe.

Explanation: A solvent is " a molecule that has the ability to dissolve other molecules". Lemon juice and sugar are solutes.

5 0
3 years ago
The general rule of thumb is that the (smaller or larger) a substance's atoms and the (stronger or weaker) the bonds, the harder
Luda [366]
Answer is: <span>The general rule of thumb is that the smaller a substance's atoms and the stronger the bonds, the harder the substance will be. 
If the distance between atoms is higher, lesser will be attraction between electrons and protons of atoms, smaller distance means stronger atoms attraction.

</span>
8 0
3 years ago
The equilibrium constant for the reaction 2NO2(g) N2O4(g) is Keq = . If a sample at equilibrium was found to contain 0.058 M NO2
Sladkaya [172]

Answer:

\boxed{3.6}

Explanation:

                  2NO₂ ⇌ N₂O₄

E/mol·L⁻¹:   0.058     0.012

K_{\text{eq}} = \dfrac{\text{[N$_{2}$O$_{4}$]}}{\text{[NO$_{2}$]$^{2}$}} = \dfrac{0.012}{0.058^{2}} = \mathbf{3.6} \\\\

\text{The $K_{\text{eq}}$ value would be $\boxed{\mathbf{3.6}}$}

7 0
3 years ago
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