Plz help ASAP plz and thx if u help

True or false Fossils teach us about the history of life

Anika [276]

True in a evolutionary sense

7 0

3 years ago

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CH4 + 202 → CO2 + 2H2O<br> How many moles O2 needed to produce 4 moles of H2O?

shepuryov [24]

<h3>Answer:</h3>

4 mol O₂

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 4 mol H₂O

[Solve] x mol O₂

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol H₂O → 2 mol O₂

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 4 \ mol \ H_2O(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})$
Multiply/Divide: $\displaystyle 4 \ mol \ O_2$

7 0

3 years ago

Read 2 more answers

Whitney's lung capacity was measured as 3.3 l at a body temperature of 37 ∘c and a pressure of 746 mmhg . how many moles of oxyg

myrzilka [38]

the ideal gas equation is PV=nRT
where P=pressure
V=Volume
n=no. of moles
R=universal gas constant
T=temperature
The universal gas constant (R) is 0.0821 L*atm/mol*K
a pressure of 746 mmhg =0.98 atm= 1 atm (approx)
T=37 degrees Celsius =37+273=310 K (convert it to Kelvin by adding 273)
V=0.7 L (only getting oxygen, get 21% of 3.3L)
Solution:
(1 atm)(0.7 L)=n(0.0821 L*atm/mol*K)(310 K)
0.7 L*atm=n(25.451 L*atm/mol)
n=0.0275 mole
Answer:
n=0.0275 mole of oxygen in the lungs.

4 0

3 years ago

Order earths layers from highest temperature to lowest temperature .

Andreyy89

Inner Core, Outer Core, Lower Mantle, Upper Mantle, Crust.

5 0

3 years ago

Given the half‑reactions and their respective standard reduction potentials 1. Cr 3 + + e − ⟶ Cr 2 + E ∘ 1 = − 0.407 V 2. Cr 2 +

Licemer1 [7]

Answer:

The standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s) is -0.744 V

Explanation:

Here we have

1. Cr³⁺ + e − ⟶ Cr²⁺ E⁰₁ = − 0.407 V

2. Cr²⁺ + 2 e − ⟶ Cr ( s ) E⁰₂ = − 0.913 V

To solve the question, we convert, the E⁰ values to ΔG as follows

ΔG₁ = n·F·E⁰₁ and ΔG₂ = n·F·E⁰₂

Where:

F = Faraday's constant in calories

n = Number of e⁻

ΔG₁ = Gibbs free energy for the first reaction

ΔG₂ = Gibbs free energy for the second half reaction

E⁰₁ = Reduction potential for the first half reaction

E⁰₂ = Reduction potential for the second half reaction

∴ ΔG₁ = 1 × F × − 0.407 V

ΔG₂ = 2 × F × − 0.913 V

ΔG₁ + ΔG₂ = F × -2.233 V which gives

ΔG = n × F × ΔE⁰ = F × -2.233 V

Where n = total number of electrons ⇒ 1·e⁻ + 2·e⁻ = 3·e⁻ = 3 electrons

We have, 3 × F × ΔE⁰ = F × -2.233 V

Which gives ΔE⁰ = -2.233 V /3 = -0.744 V.

7 0

3 years ago