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2 years ago

How does changing the amount of one substance affect a mixtures identity and a compounds identity iscience?

Chemistry

2 answers:

pychu [463]2 years ago

5 0

Explanation:

A compound is defined as follows.

Compounds are pure substance.
The atoms bonded in a compound are in specific proportion.
A compound is formed by chemical combination.
For a compound, melting point and boiling point is defined.

A mixture is defined as follows.

Mixtures are impure substance.
The bonded atoms in a mixture are in any proportion.
A mixture is formed by physical combination.
For a mixture, melting point and boiling point is not defined.

On changing the amount of one substance will affect the formation of compound as a compound requires atoms to be bonded in a specific proportion. Whereas changing the amount of one substance will not affect the formation of mixtures as atoms can be bonded in any proportion in a mixture.

KonstantinChe [14]2 years ago

3 0

According to the Law of Definite Proportions from Dalton's Atomic Theory, every compound has a fixed ratio of the mass or moles of its individual elements. This is obeyed by balancing the reactions through stoichiometric coefficients. When you increase the concentration of one substance by adding its amount, then, the other elements would also increase in order to maintain that fixed ratio.

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What is the name of the compound K3N?

alexdok [17]

The name of the compound K3N is potassium nitride (C).

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3 years ago

Read 2 more answers

Functional group and bond hybridization of vanillin

lana66690 [7]

Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.

See attached figure for the structure.

Vanillin have 3 functional groups:

1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen

2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring

3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons

Now for the hybridization we have:

The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a sp² hybridization because they are involved in double bonds.

The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a sp³ hybridization because they are involved only in single bonds.

7 0

3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

A solution has a pH of 11.8. What is the pOH?

amid [387]

Answer:

pOH is 2.2

Explanation:

The pH scale goes from 0 to 14. As such, the pH and the pOH add up to 14. Subtract your pH from 14 to get your answer.

14 - 11.8 = 2.2

5 0

3 years ago

If 25.5g of sodium thiosulphate was dissolved in 40g of distilled water at 25°C,

zhenek [66]

Answer:

Formula: Na2S2O3

we get solubility.

Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .

Solution given:

mass of sodium thiosulphate [m1]=25.5g

mass of water [m2]=40g

at temperature [t]=25°C

we have

solubility in g/dm^3 : $\frac{solute in gram}{solvent in gram} *100$

= $\frac{25.5}{40}*100$
=63.75g /litre=63.75g/dm³

solubility in g/dm^3 :63.75g/dm³

now

solubility of the solute in mol/dm^3=:63.75g/dm³/178=0.4 mol/dm³

6 0

2 years ago