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Triss [41]
3 years ago
11

The correct name for P5O2 is

Chemistry
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

phosphorous(iii) oxide

Explanation:

Hope this helps! :)

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Given the thermochemical equations below, What is the standard heat of formation of CuO(s)? 2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆
OlgaM077 [116]

-130KJ is the standard heat of formation of CuO.

Explanation:

The standard heat of formation or enthalpy change can be calculated by using the formula:

standard heat of formation of reaction = standard enthalpy of formation of product - sum of enthalpy of product formation

Data given:

Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ

2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ

CuO + Cu ⇒ Cu2O (-11.3 KJ)      ( Formation of Cu2O)

When 1 mole Cu20 undergoes combustion 1/2 moles of oxygen is consumed.

Cu20 + 1/2 02 ⇒ 2CuO (I/2 of 238.7 KJ) or 119.35 KJ

So standard heat of formation of  formation of Cu0 as:

Cu + 1/2 02 ⇒ CuO

putting the values in the equation

ΔHf = ΔH1 + ΔH2     (ΔH1 + ΔH2  enthalapy of reactants)

heat of formation = -11.3 + (-119.35)

                            = - 130.65kJ

-130.65 KJ is the heat of formation of CuO in the given reaction.

7 0
3 years ago
Read 2 more answers
Copper has two naturally occurring isotopes, ⁶³Cu (isotopic mass = 62.9296 amu) and ⁶⁵Cu (isotopic mass = 64.9278 amu). If coppe
Stells [14]

Copper has two naturally occurring isotopes, ⁶³Cu (isotopic mass = 62.9296 amu) and ⁶⁵Cu (isotopic mass = 64.9278 amu). If copper has an atomic mass of 63.546 amu,

Abundance of ⁶³Cu and ⁶⁵Cu ; 70% , 30.848%

<h3>Calculation : </h3>

Let the natural abundance of  63,  Cu isotope be x%.

The natural abundance of  

65

Cu isotope will be 100−x.

The average atomic weight is 63.546 g.

Hence, 100×63.546=63x+65(100−x)

6354.6=63x+6500−65x=6500−2x

2x=145.4

x=72.7≃70

Hence, the natural abundance of the  63Cu isotope must be approximately 70%.

For the ⁶⁵Cu calculation is same and the answer is 30.848%

<h3>What is isotope ?</h3>

Isotopes are members of a family of elements that all have the same number of protons and different numbers of neutrons. The number of protons in the nucleus determines the atomic number of an element on the periodic table.

Example : Magnesium has three natural isotopes : ²⁴Mg, ²⁵Mg, and ²⁶Mg.

<h3>What is Atomic mass ?</h3>

Atomic mass is the mass of an atom. The SI unit of mass is the kilogram, while atomic mass is often expressed in the non-SI unit dalton (synonymous with uniform atomic mass unit). 1 Da is defined as 1/12 the mass of a free carbon-12 atom at rest in the ground state.

To know more about Atomic mass please click here : brainly.com/question/338808

#SPJ4

3 0
2 years ago
Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
Free-energy change, ΔG∘, is related to cell potential, E∘, by the equationΔG∘=−nFE∘where n is the number of moles of electrons t
mart [117]

Answer:

a)\Delta G=372490 J

b)\Delta G=-568614 J

Explanation:

a) The reaction:

Mg(s) +Fe^{2+}(aq) \longrightarrow Mg^{2+}(aq) + Fe(s)

The free-energy expression:

\Delta G=-n*F*E

E=E_{red}-E_{ox]

The element wich is reduced is the Fe and the one that oxidates is the Mg:

-0.44V=E_{red}-(-2.37V)=1.93V

The electrons transfered (n) in this reaction are 2, so:

\Delta G=-2mol*96500 C/mol * 1.93 V

\Delta G=-372490 J

b) If you have values of enthalpy and enthropy you can calculate the free-energy by:

\Delta G=\Delta H - T* \Delta S

with T in Kelvin

\Delta G=-675 kJ*\frac{1000J}{kJ} - 298K*-357 J/K

\Delta G=-568614 J

7 0
3 years ago
If 100% is not passed from one trophic level to the next, where does this energy go?
Sonja [21]
D is the answer to question

6 0
3 years ago
Read 2 more answers
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