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3 years ago

Which coefficient before sodium bromide (NaBr) balances this chemical equation?

1 answer:

8 0

We see that in the left-hand side of the equation, the side of the reactants, that we have 2 moles of Na and bromine is in it's diatomic form.

Therefore, we have 2 moles of each of these elements. When we combine the sodium bromide molecule in the products, we are going to want to keep these molar amounts the same. So we are going to need to put a 2 in front of the sodium bromide in order to correctly balance this equation.

So the coefficient for sodium bromide (NaBr) in this equation is 2.

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A powder contains FeSO4⋅7H2O (molar mass=278.01 g/mol), among other components. A 3.930 g sample of the powder was dissolved in

fredd [130]

Answer:

The mass of FeSO4*7H2O in the sample is 1.21 grams

Explanation:

<u>Step 1</u>: Calculate moles of Fe2O3

moles of Fe2O3 = mass of Fe2O3 / Molar mass of Fe2O3

moles of Fe2O3 = 0.348 grams / 159.69 g/mole = 0.00218 moles

<u>Step 2</u>: Calculate moles of Fe

4 Fe + 3O2 → 2Fe2O3

For 4 moles of Fe consumed there is 2 moles of Fe2O3 produced

This means it has a ratio 2:1

So 0.00218 moles of Fe2O3 produced , there is 2*0.00218 = 0.00436 moles of Fe consumed

<u>Step 3:</u> Calculate moles of FeSO4*7H2O

Fe + H2SO4 + 7H2O → FeSO4*7H20 + H2

For 1 mole of Fe consumed there is 1 mole of FeSO4*7H2O produced

This means for 0.00436 moles there is 0.00436 moles of Fe2SO4*H2O produced

<u>Step 4:</u> Calculate the mass of FeSO4*7H2O in the sample

mass of FeSO4*7H2O = 0.00436 moles * 278.01 g/mole = 1.212 g

The mass of FeSO4*7H2O in the sample is 1.21 grams

8 0

3 years ago

Use the changes in oxidation numbers to identify which atom is oxidized, reduced, the oxidizing agent, and the reducing agent. 5

Vinil7 [7]

Answer:

Reaction A:

Hydrogen atoms in H₂ are oxidized.
Oxygen atoms in O₂ are reduced.
Hydrogen gas H₂ is the reducing agent.
Oxygen gas O₂ is the oxidizing agent.

Reaction B:

Oxygen atoms in KNO₃ are oxidized.
Nitrogen atoms in KNO₃ are reduced.
Potassium nitrate (V) KNO₃ is both the oxidizing agent and the reducing agent.

Explanation:

When an atom is oxidized, its oxidation number increases.
When an atom is reduced, its oxidation number decreases.
The oxidizing agent contains atoms that are reduced.
The reducing agent contains atoms that are oxidized.

Here are some common rules for assigning oxidation states.

Oxidation states on all atoms in a neutral compound shall add up to 0.
The average oxidation state on an atom is zero if the compound contains only atoms of that element. (E.g., the oxidation state on O in O₂ is zero.)
The oxidation state on oxygen atoms in compounds is typically -2. (Exceptions: oxygen bonded to fluorine, and peroxides.)
The oxidation state on group one metals (Li, Na, K) in compounds is typically +1.
The oxidation state on group two metals (Mg, Ca, Ba) in compounds is typically +2.
The oxidation state on H in compounds is typically +1. (Exceptions: metal hydrides where the oxidation state on H can be -1.)

For this question, only the rule about neutral compounds, oxygen, and group one metals (K in this case) are needed.

<h3>Reaction B</h3>

Oxidation states in KNO₃:

K is a group one metal. The oxidation state on K in the compound KNO₃ shall be +1.
The oxidation state on N tend to vary a lot, from -3 all the way to +5. Leave that as $x$ for now.
There's no fluorine in KNO₃. The ion NO₃⁻ stands for nitrate. There's no peroxide in that ion. The oxidation state on O in this compound shall be -2.
Let the oxidation state on N be $x$ . The oxidation state of all five atoms in the formula KNO₃ shall add up to zero. $1\times (+1) + 1 \times (x) + {\bf 3} \times (-2) = 0\\x = +5$ . As a result, the oxidation state on N in KNO₃ will be +5.

Similarly, for KNO₂:

The oxidation state on the group one metal K in KNO₂ will still be +1.
Let the oxidation state on N be $y$ .
There's no peroxide in the nitrite ion, NO₂⁻, either. The oxidation state on O in KNO₂ will still be -2.
The oxidation state on all atoms in this formula shall add up to 0. Solve for the oxidation state on N: $1\times (+1) + 1 \times (y) + {\bf 2}\times (-2) = 0\\y = +3$ . The oxidation state on N in KNO₂ will be +3.

Oxygen is the only element in O₂. As a result,

The oxidation state on O in O₂ will be 0.

$\rm\stackrel{+1}{K}\stackrel{\bf +5}{N}\stackrel{\bf -2}{O}_3 \to \stackrel{+1}{K}\stackrel{\bf+3}{N}\stackrel{\bf -2}{O}_2 + \stackrel{\bf 0}{O}_2$ .

The oxidation state on two oxygen atoms in KNO₃ increases from -2 to 0. These oxygen atoms are oxidized. KNO₃ is also the reducing agent.

The oxidation state on the nitrogen atom in KNO₃ decreases from +5 to +3. That nitrogen atom is reduced. As a result, KNO₃ is also the oxidizing agent.

<h3>Reaction A</h3>

Apply these steps to reaction A.

H₂:

Oxidation state on H: 0.

O₂:

Oxidation state on O: 0.

H₂O:

Oxidation state on H: +1.
Oxidation state on O: -2.
Double check: ${\bf 2} \times (+1) + (-2) = 0$ .

$\rm \stackrel{}{2}\; \stackrel{\bf 0}{H}_2 + \stackrel{\bf 0}{O}_2\stackrel{}{\to} \stackrel{}{2}\;\stackrel{\bf +1}{H}_2\stackrel{\bf -2}{O}$ .

The oxidation state on oxygen atoms decreases from 0 to -2. Those oxygen atoms are reduced. O₂ is thus the oxidizing agent.

The oxidation state on hydrogen atoms increases from 0 to +1. Those hydrogen atoms are oxidized. H₂ is thus the reducing agent.

4 0

3 years ago

Hi can you slove this please. ..

Crank

Answer : The balanced chemical equation will be:

(i) $2K+H_2SO_4\rightarrow K_2SO_4+H_2$

(ii) $Mg(OH)_2+Zn\rightarrow Zn(OH)_2+Mg$

Explanation :

Balanced chemical equation : It is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on product side.

Part (i):

The balanced chemical equation will be:

$2K+H_2SO_4\rightarrow K_2SO_4+H_2$

This reaction is a single displacement reaction in which most reactive element (potassium) displaces the least reactive element (hydrogen) form their solution.

Part (ii):

The balanced chemical equation will be:

$Mg(OH)_2+Zn\rightarrow Zn(OH)_2+Mg$

This reaction is a single displacement reaction in which most reactive element (zinc) displaces the least reactive element (magnesium) form their solution.

7 0

3 years ago

How many grams of Cl2 are in 1.20 x 1024 Cl atoms?

Vaselesa [24]

<h3>Answer:</h3>

70.906 g

<h3>Explanation:</h3>

We are given;

Atoms of Chlorine = 1.2 × 10^24 atoms

We are required to calculate the mass of Chlorine

We know that 1 mole of an element contains atoms equivalent to the Avogadro's number, 6.022 × 10^23.
That is , 1 mole of an element = 6.022 × 10^23 atoms

Therefore; 1 mole of Chlorine = 6.022 × 10^23 atoms

But since Chlorine gas is a molecule;

1 mole of Chlorine gas = 2 × 6.022 × 10^23 atoms

But, molar mass of Chlorine gas = 70.906 g/mol

Then;

70.906 g Of chlorine gas = 2 × 6.022 × 10^23 atoms

= 1.20 × 10^24 atoms

Thus;

For 1.2 × 10^24 atoms ;

= ( 70.906 g/mol × 1.2 × 10^24 atoms ) ÷ (1.20 × 10^24 atoms)

Therefore, 1.20 × 10^24 atoms of chlorine contains a mass of 70.906 g

5 0

3 years ago

What describes the change in oxidation states of the following reaction

miskamm [114]

Cu ions reduced to cu and Al oxidized to Al ions is the answer. copper ions is right hand side of the E.M.F cell where reduction take place and is reduced to copper metal. Al metal is at left hand side of the E.m.f cell where oxidation take place and is oxidized to Al ions

7 0

3 years ago