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VMariaS [17]
3 years ago
14

Do you think the ingredients changing into a cake

Chemistry
1 answer:
Nata [24]3 years ago
7 0
It is a chemical reaction because the baking soda or powder undergoes a chemical reaction
You might be interested in
Identify the oxidizing agent and the reducing agent in the reaction. 8H +( aq) + Cr 2O 7 2–( aq) + 3SO 3 2–( aq) → 2Cr 3+( aq) +
S_A_V [24]

Answer:

SO₃²⁻ is the reducing agent and Cr₂O₇²⁻ is the oxidizing agent.

Explanation:

Oxidation reaction:

3SO₃²⁻ (aq) + 3H₂O (l) → 3SO₄²⁻ (aq) + 6H⁺ (aq) + 6e⁻                  

Reduction reaction:

Cr₂O₇²⁻ (aq) + 14H⁺ (aq) + 6e⁻ → 2Cr ³⁺ (aq) + 7H₂O (l)

Now, adding the oxidation and the reduction reactions we get the full net reaction:

Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺ (aq) → 2Cr ³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)

Since, the S in SO₃²⁻, present in the +4 oxidation state is oxidized to +6 oxidation state in SO₄²⁻, by the loss of 2e⁻.

<u>Therefore, SO₃²⁻ is the reducing agent. </u>

And, the Cr in Cr₂O₇²⁻, present in the +6 oxidation state is getting reduced to +3 oxidation state, Cr ³⁺, by the gain of 6e⁻.

<u>Therefore, Cr₂O₇²⁻ is the oxidizing agent.</u>

3 0
2 years ago
Find the theoretical oxygen demand for the
never [62]

Answer:

a) 213.3 mg/L

b) 62.61 mg/L

c) 0.0225 mg/L

Explanation:

Theoretical oxygen demand (ThOD)is essentially the amount of oxygen required for the complete degradation of a given compound into the final oxidized products

a) Given:

Concentration of acetic acid,[CH3COOH] = 200 mg/L

CH3COOH + 2O2 \rightarrow 2CO2 + 2H2O

ThOD = \frac{mass\ O2}{mass\ CH3COOH} * conc. CH3COOH

Based on the reaction stoichiometry:

mass of CH3COOH = 60 g

mass of O2= 2(32) = 64 g

ThOD = \frac{64 g}{60 g} * 200mg/L = 213.3 mg/L

b) Given:

Concentration of ethanol, [C2H5OH] = 30 mg/L

C2H5OH + 3O2 \rightarrow 2CO2 + 3H2O

ThOD = \frac{mass\ O2}{mass\ C2H5OH} * conc. C2H5OH

Based on the reaction stoichiometry:

mass of C2H5OH = 46 g

mass of O2= 3(32) = 96 g

ThOD = \frac{96 g}{46 g} * 30mg/L = 62.61 mg/L

c) Given:

Concentration of sucrose, [C12H22O11] = 50 mg/L

C12H22O11 + 12O2 \rightarrow 12CO2 + 11H2O

ThOD = \frac{mass\ O2}{mass\ C22H22O11} * conc. C12H22O11

Based on the reaction stoichiometry:

mass of C12H22O11 = 342 g

mass of O2= 12(32) = 384 g

ThOD = \frac{384 g}{342 g} * 50mg/L = 0.0225 mg/L

7 0
3 years ago
Consider the reaction of 2.5 grams of Li (s) reacting with 0.5 grams of N2 (g) to produce Li3N (s). A) How many total grams of L
vaieri [72.5K]

Answer:

A) The amount in grams of Li₃N produced is 1.243 g

B) N₂, is the limiting reagent

The mass of the non-limiting reagent, Li, remaining after the reaction is completed is 1.757 g

Explanation:

The given parameters are;

The mass of Li(s) = 2.5 grams

The mass of N₂ (g) = 0.5 grams

The chemical equation for the reaction can be presented as follows;

6 Li (s) + N₂ (g) → 2 Li₃N

Therefore, 6 moles of Li reacts with 1 mole of N₂  to produce 2 moles of Li₃N

The molar mass of Li = 6.941 g/mol

The molar mass of N₂ = 28.0134 g/mol

The number of moles of a reactant or product, n is given by the relation;

n = Mass of substance/(Molar mass of the substance)

For lithium, Li, n = 2.5/6.941 = 0.3602 moles

For Nitrogen gas, N₂, n = 0.5/28.0134  = 0.01785 moles

A) Given that 1 mole of  N₂ to produces 2 moles of Li₃N

0.01785  moles of  N₂ will produces 2×0.01785 = 0.0357 moles of Li₃N

The molar mass of Li₃N = 34.83 g/mol

The mass of Li₃N = 34.83 g/mol × 0.0357 moles = 1.243 g

B) 6  moles of Li reacts with 1 mole of N₂ to produce 2 moles of Li₃N

0.3602 moles will reacts with 1/6×0.3602 = 0.06003 mole of N₂

Therefore, N₂, is the limiting reagent and we have;

0.01785  moles of  N₂ will react with 6×0.01785 = 0.1071  moles of Li

The number of of moles of Li left = 0.3602 - 0.1071 =0.2531 moles

The mass of lithium left = 0.2531 moles × 6.941 g/mol = 1.757 g

The mass of lithium remaining after the reaction is completed = 1.757 g.

4 0
3 years ago
Record the volume of solution that will be delivered from the pipet pictures below in units of mL
zloy xaker [14]

Answer:

2a. 3mL.

2b. 10mL.

3. 38°C.

Explanation:

2a. We read the volume under the meniscus i.e under the curve.

The volume of the solution is 3mL.

2b. The volume of the pipette 10mL and the solution is at the calibration mark. Therefore, the volume of the solution is 10mL.

3. From the diagram given, we can see that 1 line represents 1 unit.

Now a careful look at the picture shows that the temperature is 38°C.

7 0
2 years ago
A block weighing 13.15 grams has sides of 11.5 cm, 5.75 cm, and 8.44 cm. What is the density of the block in g/cm?
vitfil [10]

Answer:

B. or .0236

Explanation:

Multiply the 3 sides to get 558.095, then divide the mass and volume to get .0236

7 0
3 years ago
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