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2 years ago

Is the chemical reaction, NaHCO3 ⇒ Na2CO3 + H2O + CO2, an example of combustion or an example of decomposition?

Chemistry

2 answers:

Lelu [443]2 years ago

6 0

Answer:

This chemical reaction is a decomposition reaction this is because this reaction breaks up the chemical species breaking up into similar parts.

hope I helped x

Tema [17]2 years ago

6 0

Answer:

it's decomposition

Explanation:

because it breaks up the chemical into smaller parts

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what is the balanced equation for the reaction of solid magnesium metal with oxygen gas to produce solid magnesium oxide

GaryK [48]

Answer:

2Mg + O₂ ⟶ 2MgO

Explanation:

Step 1. Start with the most complicated-looking formula (O₂?).

Put a 1 in front of it.

Mg + 1O₂ ⟶ MgO

Step 2. Balance O.

We have fixed 2 O on the left. We need 2O on the right. Put a 2 in front of MgO.

Mg + 1O₂ ⟶ 2MgO

Step 3. Balance Mg.

We have fixed 2 Mg on the right-hand side. We need 2 Mg atoms on the left. Put a 2 in front of Mg.

2Mg + 1O₂ ⟶ 2MgO

Every formula now has a coefficient. The equation should be balanced. Let’s check.

Atom On the left On the right

Mg 2 2

O 2 2

All atoms are balanced.

The balanced equation is

2Mg + O₂ ⟶ 2MgO

4 0

3 years ago

Write the type of reaction for: Ca(OH)2 + Al2(SO4)3 \rightarrow CaSO4 + Al(OH)3

antiseptic1488 [7]

This is an ionic precipitation reaction

8 0

3 years ago

A 1.30M solution of BaCl2 has a density of 1.230 g/ml. a) What is the mole fraction of BaCl2 in this solution?

Elodia [21]

Answer: The mole fraction of barium chloride in the solution is 0.024

Explanation:

We are given:

Molarity of barium chloride solution = 1.30 M

This means that 1.30 moles of barium chloride is present in 1 L or 1000 mL of solution.

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.230 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.230g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.230g/mL\times 1000mL)=1230g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Moles of barium chloride = 1.30 moles

Molar mass of barium chloride = 208 g/mol

Putting values in equation 1, we get:

$1.30mol=\frac{\text{Mass of barium chloride}}{208g/mol}\\\\\text{Mass of barium chloride}=(1.30mol\times 208g/mol)=270.4g$

Mass of water = Mass of solution - Mass of barium chloride

Mass of water = 1230 - 270.4 = 959.6 g

Calculating the moles of water:

Given mass of water = 959.6 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{959.6g}{18g/mol}\\\\\text{Moles of water}=53.31mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

For barium chloride:

$\chi_{\text{(barium chloride)}}=\frac{n_{\text{(barium chloride)}}}{n_{\text{(water)}}+n_{\text{(barium chloride)}}}$

$\chi_{\text{(barium chloride)}}=\frac{1.30}{1.30+53.31}\\\\\chi_{\text{(barium chloride)}}=0.024$

Hence, the mole fraction of barium chloride in the solution is 0.024

6 0

4 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago

If a man is 50 years old how many times would he go around the sun on mecury

ValentinkaMS [17]

It only takes mercury 88 days to go around the sun, and if the man is 50 years old he would be 18,250 days old, so he would've gone around the sun 207.39 times.

5 0

3 years ago