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juin [17]
3 years ago
12

What is the mass of calcium in 2.3•10^23 molecules of calcium phosphate

Chemistry
1 answer:
Katena32 [7]3 years ago
3 0

Explanation:

N(Ca)=3×N(Ca₃(PO₄)₂)=

=3×2.3×10²³=6.9×10²³

6.02×10²³→40g

6.9×10²³→Xg

m(Ca)=X≈45.85g

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Many power plants produce energy by burning carbon-based fuels, which also produces carbon dioxide. Carbon dioxide is a greenhou
RUDIKE [14]

Answer:

a) 2.541 mol/MJ;

b) 1.124 mol/MJ;

c) 0.4354 mol/MJ;

d) 0.1835 mol/MJ

Explanation:

The enthalpy of formation (ΔH°f) is the enthalpy of a reaction to form a compound by its constituents. For CO₂, ΔH°f = - 393.5 kJ/mol.

The enthalpy of a reaction is the sum of the enthalpy of the products (each one multiplied by the number of moles) less the sum of the enthalpy of the reactants (each one multiplied by the number of moles). The ΔH°f for simple substances (with one atom) is 0. The combustion is the reaction between the fuel and the oxygen.

a) The combution reaction is:

C(s) + O₂(g) → CO₂(g)

ΔH°rxn = -393.5 kJ/mol = -393.5x10⁻³ MJ/mol

Number of moles per MJ released: 1/|ΔH°rxn|

n = 1/(393.5x10⁻³) = 2.541 mol/MJ

b) The combustion reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

H₂O is in the liquid state because it's at 1 atm and 25ºC.

ΔH°f, H₂O(l) = -285.3 kJ/mol

ΔH°f, O₂(g) = 0

ΔH°f, CH₄(g) = -74.8 kJ/mol

ΔH°rxn = [2*(-285.3 ) + 1*(-393.5)] - [1*(-74.8)]

ΔH°rxn = -889.3 kJ/mol = -889.3x10⁻³ MJ/mol

n = 1/889.3x10⁻³ = 1.124 mol/MJ

c) C₃H₈(g) + 10O₂(g) → 3CO₂(g) + 4H₂O(l)

ΔH°f,C₃H₈(g) = -25.2 kJ/mol

ΔH°rxn = [4*(-285.3) + 3*(-393.5)] - [1*(-25.2)]

ΔH°rxn = -2,296.5 kJ/mol = -2.2965 MJ/mol

n = 1/2.2965 = 0.4354 mol/MJ

d) C₈H₁₈(l) + (25/2)O₂(g) → 8CO₂(g) + 9H₂O(l)

ΔH°f, C₈H₁₈(l) = -250.1 kJ/mol

ΔH°rxn = [9*(-283.5) + 8*(-393.5)] - [1*(-250.1)]

ΔH°rxn = -5,449.4 kJ/mol = -5.4494 MJ/mol

n = 1/5.4494 = 0.1835 mol/MJ

4 0
3 years ago
17)<br> How many grams are in 0.02 moles of beryllium iodide, Bel2?
Alinara [238K]

Answer:

beryllium iodide has a molar mass of 262.821 g mol−1 , which means that 1 mole of beryllium iodide has a mass of 262.821 g . To find the mass of 0.02 moles of beryllium iodide, simply multiply the number of moles by the molar mass in conversion factor form.

Explanation:

5 0
3 years ago
Read 2 more answers
Hello Friends Can Yall Plz Help Me For Brainliest Don’t Comments No B.S Yu Will Get Cursed out show ya work
VARVARA [1.3K]

Answer:

it is b

Explanation:

mid ocean ridge diverges meaning it moving in two different direction horizontally - left to right

4 0
3 years ago
Identify the products of this single-replacement reaction between sulfuric acid and aluminum. Then balance the equation. (2 poin
jeka94

The products of this reaction between aluminum and sulfuric acid are two: hydrogen and aluminum sulfate.

<h3>What are the products in a reaction?</h3>

This concept refers to the substances obtained at the end of the reaction.

<h3>What does it mean to balance an equation?</h3>

It means to make sure there are the same molecules in the reactants and products.

<h3>What are the products in this reaction?</h3>
  • Al + H2SO4 = Al2 (SO4)3 + H2  
  • Al2 (SO4)3 = Aluminium sulfate
  • H = Hydrogen

<h3>What is the balanced equation?</h3>

2Al + 3H2SO4 = Al2 (SO4)3 + 3H2

Learn more about chemical reaction in: brainly.com/question/3461108

7 0
2 years ago
What is the mass of 1.58 moles of CH4
HACTEHA [7]
<h3>Answer:</h3>

25.4 g CH₄

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<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>

1.58 mol CH₄

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

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 1.58 \ mol \ CH_4(\frac{16.05 \ g \ CH_4}{1 \ mol \ CH_4})
  2. Multiply/Divide:                 \displaystyle 25.359 \ g \ CH_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

25.359 g CH₄ ≈ 25.4 g CH₄

4 0
3 years ago
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