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3 years ago

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Match each term to its description. Match Term Definition Excess reactant A) Reactant that can produce more of the product Limit

ing reactant B) Reactant that can produce a lesser amount of the product Theoretical yield C) Amount of product predicted to be produced by the given reactants

1 answer:

faust18 [17]3 years ago

5 0

Answer:

Explanation:

A) Reactant that can produce more of the product

Excess reactant:

In a given reaction, the reactant that is in excess supply is the excess reactant. If the amount of the excess reactant is match, more of the product will be produced.

B) Reactant that can produce a lesser amount of the product

Limiting reactant

The limiting reactant restricts the progress of the reaction. It determines the amount of product that can be formed.

C) Amount of product predicted to be produced by the given reactants

Theoretical yield

For a given amount of reactants, the theoretical yield determines the amount of products that can be produced.

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The atomic mass of oxygen is 15.999 amu whatbdoes the value equal

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Answer:

The sum of the protons and neutrons.

Explanation:

protons + neutrons = atomic mass

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3 years ago

What is one way that electric charges are different from magnetic poles?

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3 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

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3 years ago

G vinegar is a solution of acetic acid in water. if a 165 ml bottle of distilled vinegar contains 28.2 ml of acetic acid, what i

Delicious77 [7]

<span>Volume Percent (volume/volume%) is defined to be the ratio of the volume of solute to the volume of solution times 100%. The acetic acid is the solute volume. The bottle of vinegar is the solution volume. Acetic acid (28.2ml) / Vinegar(165ml) x 100% = 17.1%</span>

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3 years ago

Write a balanced chemical equation for the standard formation reaction of liquid acetic acid hch3co2.

Afina-wow [57]

The balanced chemical equation for the standard formation reaction of liquid acetic acid is given as ,

$2C(gr) +2H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The reaction that form the products from their elements in their standard state is called formation of reaction .The acetic acid consist C , H , and O , So, determine their standard state . Carbon is graphite at 25°C and 1 atm , whereas hydrogen and oxygen are diatomic gases . Hence , we start with unbalanced reaction.

$C(gr) +H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The balanced chemical equation for the standard formation reaction of liquid acetic acid as,

$2C(gr) +2H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The combustion of liquid acetic acid is given as,

$CH_{3} COOH(l) + 2O(g)$ → $2CO_{2}((g) +2H_{2} O(l)$ ΔH =-873

learn more about balancing chemical equation

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2 years ago