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Fiesta28 [93]
3 years ago
9

How do I find percent mass composition

Chemistry
1 answer:
castortr0y [4]3 years ago
5 0

Answer:

1. Find the molar mass of all the elements in the compound in grams per mole.

2. Find the molecular mass of the entire compound.

3. Divide the component's molar mass by the entire molecular mass.

4. You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.

Explanation:

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Calculate the standard potential for the following galvanic cell: Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
Nataly [62]

Answer:

E° = 1.24 V

Explanation:

Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)

According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:

Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻

Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an

E° = 0.80 V - (-0.44 V) = 1.24 V

6 0
3 years ago
For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
Aneli [31]
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
8 0
3 years ago
An energy of 6.8 x 10^-19 J/atom is required to cause an aluminum atom on a metal surface to lose an electron.
Nana76 [90]

Wavelength of the light is 2.9 × 10⁻⁷ m.

<u>Explanation:</u>

Planck - Einstein equation shows the relationship between the energy of a photon and its frequency, and they are directly proportional to each other and  it is given by the equation as E = hν,

where E is the energy of the photon

h is the Planck's constant = 6.626 × 10⁻³⁴ J s

ν is the frequency

From the above equation, we can find the frequency by rearranging the equation as,

ν = $ \frac{E}{h} = $ \frac{6.8 \times 10^{-19}}{6.626\times10^{-34}} = 1.03\times10^{15} s^{-1}

Now the frequency and the wavelength are in inverse relationship with each other.

ν × λ = c

It can be rearranged to get λ as,

λ = c / ν

 = $\frac{3\times 10^{8} ms^{-1}}{1.03\times10^{15}s^{-1}} = 2.9\times 10^{-7} m

So wavelength is 2.9 × 10⁻⁷ m.

6 0
3 years ago
Thermal energy is the energy an object has due to the _____ of the particles
MrRissso [65]

Answer: Vibråtory movement.

Explanation: when particles bounce against each other the friction creates thermal energy. Think about what happens when you rub your hands together and they get warmer, that the friction between your hands making thermal energy.

8 0
3 years ago
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
3 years ago
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