All categories

3 years ago

Which is a mixture?

Chemistry

2 answers:

Rama09 [41]3 years ago

5 0

Potassium and chlorine

Eva8 [605]3 years ago

4 0

Answer: Option (d) is the correct answer.

Explanation:

When two different atoms with similar composition are combined together by a chemical reaction then the product formed is known as compound.

For example, Potassium chloride is a compound as it has 1:1 composition, that is, one potassium atom is combined with one chlorine atom.

On the other hand, potassium metal is an element because it consists of only atoms of potassium.

Whereas when two different elements combine together and have different composition then it results in the formation of a mixture.

For example, potassium and chloride is a mixture as we cannot tell its composition without conducting an experiment.

Thus, we can conclude that out of the given options potassium and chlorine is a mixture.

You might be interested in

A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74

stellarik [79]

Explanation:

Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

= 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product $(CH3COOH)^{2}$ gas formed are calculated as

0.010 mol CH3COOH * 1 mol $(CH3COOH)^{2}$ / 2 mol CH3COOH

= 0.005 mol $(CH3COOH)^{2}$

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^{2}$

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

= 0.74 atm * 0.015 mol / 0.020 mol

= 0.555 atm

4 0

3 years ago

How many moles of carbonate are there in sodium carbonate

vaieri [72.5K]

There are 0.566 moles of carbonate in sodium carbonate.

<h3>CALCULATE MOLES:</h3>

The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.

no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3

Molar mass of Na2CO3 = 23(2) + 12 + 16(3)

= 46 + 12 + 48 = 106g/mol

mass of CO3 = 12 + 48 = 60g

no. of moles of CO3 = 60/106

no. of moles of CO3 = 0.566mol

Therefore, there are 0.566 moles of carbonate in sodium carbonate.

Learn more about number of moles at: brainly.com/question/1542846

5 0

2 years ago

Read 2 more answers

calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate

kirza4 [7]

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

brainly.com/question/9240031

#SPJ4

4 0

1 year ago

Is peridotite a mineral,a rock,or neither

mamaluj [8]

Answer: a rock

Explanation:

4 0

3 years ago

Read 2 more answers

What is the reaction quotient, Q, for this system when [N2] = 2.00 M, [H2] = 2.00 M, and [NH3] = 1.00 M at 472°C?

cupoosta [38]

Answer : The value of reaction quotient, Q is 0.0625.

Solution : Given,

Concentration of $N_2$ = 2.00 M

Concentration of $H_2$ = 2.00 M

Concentration of $NH_3$ = 1.00 M

Reaction quotient : It is defined as a concentration of a chemical species involved in the chemical reaction.

The balanced equilibrium reaction is,

$N_2+3H_2\rightleftharpoons 2NH_3$

The expression of reaction quotient for this reaction is,

$Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}$

Now put all the given values in this expression, we get

$Q=\frac{(1.00)^2}{(2.00)^1(2.00)^3}=0.0625$

Therefore, the value of reaction quotient, Q is 0.0625.

3 0

3 years ago