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djyliett [7]
3 years ago
9

. Consider the following half-reactions:

Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
7 0

d. Fe(s) and Al(s)

<h3>Further explanation</h3>

In the redox reaction, it is also known  

Reducing agents are substances that experience oxidation  

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series  

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au  </em>

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent  

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction  

The electrodes which are easier to reduce than hydrogen (H), have E cells = +

The electrodes which are easier to oxidize than hydrogen have a sign E cell = -

So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)

The metal  : d. Fe(s) and Al(s)

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A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What i
In-s [12.5K]

Answer:

81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>

<em />

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

Actual yield (0.366g) / Theoretical yield * 100

<em>Moles CaCl₂ = Moles CaCO₃:</em>

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield

3 0
3 years ago
How much water (H2O ) would form if 4.04 g of hydrogen (H2) reacted with 31.98 g of oxygen (O2 )?
Norma-Jean [14]

Answer:

Mass = 36 g

Explanation:

Given data:

Mass of water formed = ?

Mass of hydrogen = 4.04 g

Mass of oxygen = 31.98 g

Solution:

Chemical equation:

2H₂ + O₂   →   2H₂O

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 4.04 g/ 2 g/mol

Number of moles = 2.02 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 31.98 g/ 32 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of water with hydrogen and oxygen.

                O₂         :         H₂O

                 1           :           2

                H₂         :         H₂O

                 2          :          2

               2.02       :      2.02

Number of moles of water formed by oxygen are less thus oxygen will limiting reactant.

Mass of water:

Mass = number of moles × molar mass

Mass = 2 mol × 18 g/mol

Mass = 36 g

8 0
2 years ago
What are 4 similarities of Metals, Nonmetals, and Metalloids?
igor_vitrenko [27]
The answer is           


e o i and a



not really sure forgive me if im wrong


7 0
3 years ago
The 225 g FeCl2 is about 1.8 moles.
Lyrx [107]

Answer:

4 \frac{mol}{l}

Explanation:

From the question, we have been asked to find the molarity of FeCl2 having a volume of 450 mL,

We have been provided with 225 g which is proportional to 1.8 moles.

We know that molarity of any solution should be in mol/L.

1 mole contained in 1 L means it has a molarity of 1 mol/L

Let's convert 450 mL to Litres which is,

\frac{450}{1000} litres

= 0.450 L

Thus,

1 mole is contained in 1L

x moles are contained in 0.450 L

Hence,

x mole/molarity = {1 mole x 1 L}/{0.450 L}

= 4 mol/L

Therefore 4 mol/L is the molar concentration.

5 0
2 years ago
The chemical formula for magnesium oxide is MgO. A chemist determined by measurements that 0.030 moles of magnesium oxide partic
tamaranim1 [39]

Answer:

1.209g of MgO participates

Explanation:

In this problem, we have 0.030 moles of MgO that participates in a particular reaction.

And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.

To convert moles to grams we need to use molar mass of the compound:

<em>1 atom of Mg has a molar mass of 24.3g/mol</em>

<em>1 atom of O has a molar mass of 16g/mol</em>

<em />

That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol

And mass of 0.030 moles of MgO is:

0.030 moles MgO * (40.3g/mol) =

<h3>1.209g of MgO participates</h3>
3 0
3 years ago
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