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murzikaleks [220]
3 years ago
11

Given the following reactions and subsequent ∆H values, what is the ∆H for the reaction below? 4CO2(g) + 2H2O(g) 2C2H2(g) + 5O2(

g)
Chemistry
1 answer:
melisa1 [442]3 years ago
7 0
The problem ask on which of the following reactions and subsequent values and the value is given on you question. Base on your data and also to my formulation, solution and further simplification about the problem and the given values, i came up with an answer of <span>∆Hrxn = 5∆Hf(O2(g)) + 2∆Hf(C2H2(g)) - 2∆Hf(H2O(g)) - 4∆Hf(CO2(g)).</span>
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KCl and KBr are both ionic solids. A mixture of KCl and KBr has a mass of 3.595 g. When this mixture is heated in the presence o
monitta

Answer:

The percentage (by mass) of KBr in the original mixture was 33.1%.

Explanation:

The mixture of KCl and KBr has a mass of 3.595g, thus the sum of the moles of KCl (<em>x</em>) multiplied by it molar mass (74.5g/mol) and the moles of KBr (<em>y</em>) multiplied by it molar mass (119g/mol) is the total mass of the mixture:

x.74.5g/mol + y.119g/mol = 3.595g

Also, after the conversion of KBr into KCl, the total mass of 3.129 g is only from KCl moles, hence

\frac{3.129g}{74.5g/mol} = 0.042 moles

But the 0.042 moles came from the originals KCl and KBr moles, thus

x + y = 0.042moles

Now it is possible to propose a system of equations:

x.74.5g/mol + y.119g/mol = 3.595g

x + y = 0.042moles

Solving the system of equations,

x=0.032moles\\y=0.010 moles

0.010 moles of KBr multiplied it molar mass is

0.010molesx119g/mol = 1.19g

Therefore, the percentage (by mass) of KBr in the original mixture was:

\frac{1.19g}{3.595g}x100% = 33.1%%

4 0
3 years ago
an atom contains 32 protons and has a mass number of approximately 72 amu what is the chemical symbol for the atom
olga nikolaevna [1]

I belive it would be Hf (hafanium).

5 0
3 years ago
Read 2 more answers
How many kilograms of iron can be obtained from 100 kilograms of Fe203
ANTONII [103]

Answer:

754

Explanation:

5 0
2 years ago
A company is testing drinking water and wants to ensure that ca content is below 155 ppm. What is the maximum amount of Ca that
BabaBlast [244]

Answer:

138 mg

Explanation:

A company is testing drinking water and wants to ensure that Ca content is below 155 ppm (= 155 mg/kg), that is, <em>155 milligrams of calcium per kilogram of drinking water</em>. We need to find the maximum amount of calcium in 890 g of drinking water.

Step 1: Convert the mass of drinking water to kilograms.

We will use the relation 1 kg = 1000 g.

890g \times \frac{1kg}{1000g} =0.890kg

Step 2: Calculate the maximum amount of calcium in 0.890 kg of drinking water

0.890gH_2O \times \frac{155mgCa}{1kgH_2O} = 138mgCa

7 0
3 years ago
CH4 + O2 → CO2 + H2O
Scilla [17]

Answer:

9.8 × 10²⁴ molecules H₂O

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Organic</u>

  • Naming carbons

<u>Stoichiometry</u>

  • Analyzing reaction rxn
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O

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

[Given] 130 g CH₄

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

Avogadro's Number

[RxN] 1 mol CH₄ → 2 mol H₂O

[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: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                   \displaystyle 130 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ H_2O}{1 \ mol \ CH_4})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 9.75526 \cdot 10^{24} \ molecules \ H_2O

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O

8 0
3 years ago
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