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3 years ago

C2H4O2 + NaC2H3O2 + H2O + CO2 list the reactants in this chemical reaction

Chemistry

1 answer:

lisabon 2012 [21]3 years ago

7 0

Answer:

C₂H₄O₂ and NaC₂H₃O₂ are reactants.

Explanation:

Word equation:

Acetic acid + sodium acetate → sodium diacetate

Chemical equation:

C₂H₄O₂ + NaC₂H₃O₂ → C₄H₇NaO₄

This is a synthesis reaction in which simple reactants combine to form complex product.

This is also balanced chemical equation because there are equal number of atoms of all elements on both side of equation. Thus it follow the law of conservation of mass.

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

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Sort the compounds: conjugated or non conjugated. -but-2-enoic acid -butanoic acid -hex-4-en-3-one -hexan-3-one

Oliga [24]

By definition of conjugated and non-conjugated acids, Conjugate acids are a type of acid that gains a proton in solution in response to a base that has happily accepted a proton.

This how the compounds are to be sorted.
-butanoic acid
-but-2-enoic acid
-hexan-3-one
-hex-4-en-3-one

3 0

3 years ago

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Questions

Ulleksa [173]

Answers:

See below

Step-by-step explanation:

1. Most food energy

(a) Pringles

Heat from Pringles + heat absorbed by water = 0

m₁ΔH + m₂CΔT = 0

1.984ΔH + 100 × 4.184 × 18 = 0

1.984ΔH + 7530 = 0

ΔH = -7530/1.984 = -3800 J/g

(b) Cheetos

0.884ΔH + 418.4 × 13 = 0

ΔH = -5400/0.884 = -6200 J/g

Cheetos give you more food energy per gram.

(c) Snickers

Food energy = 215 Cal/28 g × 4184 J/1 Cal = 32 000 J/g

The food energy from Cheetos is much less than that from a Snickers bar

2. Experimental uncertainty

The experimental values are almost certainly too low.

Your burning food is heating up the air around it, so much of the heat of combustion is lost to the atmosphere.

3. Percent efficiency

Experimental food energy = 3800 J/g

Actual food energy = 150 Cal/28 g × 4184 J/1 Cal = 22 000 J/g

% Efficiency = Experimental value/Actual value × 100 %

= 3800/22 000 × 100 %

= 17 %

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3 years ago

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Galina-37 [17]

Can we see the graph

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3 years ago

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For each of the reactions, calculate the mass (in grams) of the product formed when 15.12g of the underlined reactant completely

Ludmilka [50]

Answer:

Part A : amount of product (KCl) = 28.88 g

Part B : amount of product (KBr) = 46.13 g

Part C : amount of product (Cr₂O₃) = 17.3 g

Part D: amount of product (SrO) = 35.76 g

Explanation:

Part A:

Data Given:

Reaction :

2K(s) + Cl₂(g) --------> 2KCl

Amount of underline Reactant (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is chlorine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

2K(s) + Cl₂(g) --------> 2KCl

2mol 1mol 2mol

it shows that

2 mole of K give 2 mole of KCl

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KCl = (39 + 35.5)

Molar mass of KCl = 74.5 g/mol

So the look again to reaction in terms of grams

2K(s) + Cl₂(g) --------> 2KCl

2mole (39 g/mol) 2mole (74.5 g/mol)

78 g 149 g

Apply the Unity formula

78 g of Potassium ≅ 149 g of KCl

Then

15.12 g of Potassium ≅ how many g of Product (KCl)

By doing cross multiplication

X g of Product (KCl) = 149 g of KCl x 15.12 g of K / 78 g of K

X g of Product (KCl) = 28.88 g

So the amount of product (KCl) = 28.88 g

_________________________________________

Part B:

Data Given:

Reaction :

2K(s) + Br₂(g) --------> 2KBr

Amount of underline Reactant (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is Bromine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

2K(s) + Br₂(g) --------> 2KBr

2mol 1mol 2mol

it shows that

2 mole of K give 2 mole of KBr

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KBr = (39 + 80)

Molar mass of KBr = 119 g/mol

So, look again to reaction in terms of grams

2K(s) + Br₂(g) --------> 2KBr

2mole (39 g/mol) 2mole (119 g/mol)

78 g 238 g

Apply the Unity formula

78 g of Potassium ≅ 238 g of KBr

Then

15.12 g of Potassium ≅ how many g of Product (KBr)

By doing cross multiplication

X g of Product (KBr) = 238 g of KBr x 15.12 g of K / 78 g of K

X g of Product (KBr) = 46.13 g

So the amount of product (KBr) = 46.13 g

__________________________________________

Part C:

Data Given:

Reaction :

4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

Amount of underline Reactant (Cr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Chromium (Cr) is 15.12g and other reactant that is Oxygen is more than enough so the Cr is limiting reagent.

So, amount of product depend on the amount of Chromium (Cr)

Now Look at Given Reaction:

4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

4mol 3mol 2mol

it shows that

4 mole of Cr give 2 mole of Cr₂O₃

if we represent mole in grams

Then

Molar mass of Cr = 52 g/mol

Molar mass of 2Cr₂O₃ = 2 [2 (52) + 3(16) ] = 2 (104+ 48)

Molar mass of 2Cr₂O₃ = 304 g/mol

So, look again to reaction in terms of grams

4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

4 mol (52 g/mol) 2 mole (304 g/mol)

208 g 608 g

Apply the Unity formula

208 g of Chromium ≅ 608 g of Cr₂O₃

Then

15.12 g of Chromium ≅ how many g of Product (Cr₂O₃)

By doing cross multiplication

X g of Product (Cr₂O₃) = 238 g of Cr₂O₃ x 15.12 g of Cr / 208 g of Cr

X g of Product (Cr₂O₃) = 17.3 g

So the amount of product (Cr₂O₃) = 17.3 g

________________________________________

Part D:

Data Given:

Reaction :

2Sr(s) + O₂(g) --------> 2SrO(s)

Amount of underline Reactant (Sr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Strontium (Sr) is 15.12g and other reactant that is Oxygen is more than enough so the Sr is limiting reagent.

So, amount of product depend on the amount of Strontium (Sr)

Now Look at Given Reaction:

2Sr(s) + O₂(g) --------> 2SrO(s)

2mol 1mol 2mol

it shows that

2 mole of Sr give 2 mole of SrO

if we represent mole in grams

Then

Molar mass of Sr = 87.6 g/mol

Molar mass of 2SrO = 2 [87.6 + 16] = 2 (103.6)

Molar mass of 2SrO = 207.2 g/mol

So, look again to reaction in terms of grams

2Sr(s) + O₂(g) --------> 2SrO(s)

2 mol ( 87.6 g/mol) 2 mole (207.2 g/mol)

175.2 g 414.4 g

Apply the Unity formula

175.2 g of Strontium ≅ 414.4 g of SrO

Then

15.12 g of Strontium ≅ how many g of Product (SrO)

By doing cross multiplication

X g of Product (SrO) = 414.4 g of SrO x 15.12 g of Sr / 175.2 g of Sr

X g of Product (SrO) = 35.76 g

So the amount of product (SrO) = 35.76 g

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