State Hess' law of constant heat summation.
(b) Calculate the enthalpy of formation of CH4 from the following data:
i) C(s) + O2(g) → CO2(g); ∆H = -393.7 kJ/mol
ii) H2(g) + 1⁄2 O2(g) → H2O(l); ∆H = -285.8 kJ/mol
iii) CH4(g) + 2 O2(g)→ CO2(g) + 2H2O(l); ∆H = -890.4 kJ/mol

Explanation:

Phosphorus (P) because of the 5 valence electrons total, 3 of them are in the 3p sublevel, and according to Hund's rule, they "single-fill" each orbital first.

Answer:

you can see the empirical formula at the pic

The empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

Empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a compound.

(a) 1.245 g Ni : 5.381 g I

Mole of Ni ; Mole of I = 1.245/59 : 5.381/127 = 0.02 : 0.04 = 1:2

So the formula is NiI2

(b) 2.677 g Ba : 3.115 g Br

Mole of Ba : Mole of Br = 2.677/137 : 3.115/60 = 0.019 : 0.038

= 0.02 : 0.04 = 1:2

So the formula is BaBr2

(c) 2.128 g Be : 7.557 g S

Mole of Be : Mole of S = 2.128/9 : 7.557/32 = 0.2 : 0.2 = 1:1

So the formula is BeS

Thus, empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

To learn more about empirical formula, refer to the link below:

https://brainly.com/question/11588623

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Answer:

true

Explanation:

I'm not sure why cause I dont know how to explain but it's TRUE

Answer:

True  

Explanation:

The particles moving faster in a substance the hotter it gets.

Answer:

during sublimation

Explanation:

just took the test

Answer:

Explanation:

Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .

Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .

Answer:

Explanation:

In my opinion the answer should be SO2

Answer:

a should be answer i think.

Answer:

3.65 mol O₂

Explanation:

Step 1: RxN

2H₂O₂ → 2H₂O + O₂

Step 2: Define

Given - 7.30 mol H₂O₂

Solve - x mol O₂

Step 3: Stoichiometry

[tex]7.30 \hspace{3} mol \hspace{3} H_2O_2(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} H_2O_2} )[/tex] = 3.65 mol O₂

Answer:

atom

Explanation:

the atom is the smallest form.

Answer:

Explanation:

Kb of  (CH₃)₃N is 7.4 x 10⁻⁵

initial concentration of (CH₃)₃N   a   is .48 M

(CH₃)₃N    +   H₂O =  (CH₃)₃NH⁺  +  OH⁻

a - x                                     x               x  

x² / (a - x )  = Kb

x is far less than a so a - x can be replaced by a .

x² / a   = Kb

x²  = a x Kb = .48 x 7.4 x 10⁻⁵ = 3.55 x 10⁻⁵ = 35.5 x 10⁻⁶

x = 5.96 x 10⁻³

pOH = - log ( 5.96 x 10⁻³ )

= 3 - log 5.96

= 3 - .775

= 2.225

Answer:

A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.

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Answer:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Explanation:

Hello.

In this case, for the undergoing chemical reaction:

[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]

We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

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Answer:

[tex]Total = 50.6\ moles[/tex]

Explanation:

Given

[tex]Propane = C_3H_8[/tex]

Represent Carbon with C and Hydrogen with H

[tex]C = 13.8[/tex]

Required

Determine the total moles

First, we need to represent propane as a ratio

[tex]C_3H_8[/tex] implies

[tex]C:H = 3:8[/tex]

So, we're to first solve for H when [tex]C = 13.8[/tex]

Substitute 13.8 for C

[tex]13.8 : H = 3 : 8[/tex]

Convert to fraction

[tex]\frac{13.8}{H} = \frac{3}{8}[/tex]

Cross Multiply

[tex]3 * H = 13.8 * 8[/tex]

[tex]3 H = 110.4[/tex]

Solve for H

[tex]H = 110.4/3[/tex]

[tex]H = 36.8[/tex]

So, when

[tex]C = 13.8[/tex]

[tex]H = 36.8[/tex]

[tex]Total = C + H[/tex]

[tex]Total = 13.8 + 36.8[/tex]

[tex]Total = 50.6\ moles[/tex]

Answer:

266Lr

Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.

Explanation:

hopefully that helps you