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3 years ago

Iron(III) chloride (aq) + silver nitrate (aq) –

Chemistry

2 answers:

Katarina [22]3 years ago

5 0

Answer:

Xa-xa-xa

Explanation:

Kisachek [45]3 years ago

4 0

Reaction:
FeCl3(aq) + 3AgNO3(aq) → Fe(NO3)3(aq) + 3AgCl(s)

Net ionic:
Ag + (aq) + Cl − (aq) → AgCl(s)

Explanation:
The evidence that a double replacement reaction has happened is the formation of a solid, liquid or gas from aqueous solutions. I’m this case case silver chloride is a precipitate.

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4 years ago

4. Suppose 8.00 g of CH4 is allowed to burn in the presence of 16.00 g of oxygen. CH4(g)+2O2(g)-->CO2(g)+2H2O(g)

umka21 [38]

<h3>Answer:</h3>

No masses of CH₄ and O₂ remained after the reaction, while 22.005 g of CO₂ and 18.02 g of H₂O remained

<h3>Explanation:</h3>

The combustion of methane is given by the reaction;

CH₄(g)+2O₂(g) → CO₂(g)+2H₂O(g)

We are given, 8 g of CH₄ and 16.00 g of O₂

Required to determine the mass of CH₄, O₂, CO₂ and H₂O that remained after the complete reaction.

<h3>Step 1: Moles of CH₄ and O₂ in the mass given </h3>

Moles = mass ÷ molar mass

Molar mass of CH₄ = 16.04 g/mol

Moles of CH₄ = 8.00 g ÷ 16.04 g/mol

= 0.498 moles

= 0.5 moles

Molar mass of O₂ = 16.0 g/mol

Moles of O₂ = 16.00 g ÷ 16.00 g/mol

= 1 mole

From the reaction, 1 mole of CH₄ reacts with 2 moles of O₂

CH₄ is the limiting reactant since it is way less than the amount of O₂

Therefore, 0.5 moles of CH₄ will react with 1 mole of oxygen.

This means there will be no amount of O₂ and CH₄ that remains.

<h3>Step 2: Moles of CO₂ and H₂O that were produced.</h3>

From the reaction 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O.

Therefore,

In our case, 0.5 moles of CH₄ will react with 1 mole of O₂ to produce 0.5 moles of CO₂ and 1 mole of H₂O.

<h3>Step 3: Mass of CO₂ and H₂O produced </h3>

Mass = Moles × Molar mass

Molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.5 mol × 44.01 g/mol

= 22.005 g

Molar mass of H₂O = 18.02 g/mol

Moles of H₂O = 1 mole × 18.02 g/mol

= 18.02 g

Therefore, no masses of CH₄ and O₂ remained after the reaction, 22.005 g of CO₂ and 18.02 g of H₂O remained

3 0

4 years ago

Read 2 more answers

1) How many atoms are in 0.54 moles of Cu? show work

mario62 [17]

<h3>Answer:</h3>

3.3 × 10²³ atoms Cu

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 0.54 moles Cu

[Solve] atoms Cu

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 0.54 \ mol \ Cu(\frac{6.022 \cdot 10^{23} \ atoms \ Cu}{1 \ mol \ Cu})$
[DA] Multiply [Cancel out units]: $\displaystyle 3.25188 \cdot 10^{23} \ atoms \ Cu$

<u>Step 4: Check</u>

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

3.25188 × 10²³ atoms Cu ≈ 3.3 × 10²³ atoms Cu

7 0

3 years ago