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3 years ago

Which of the following quantities contains the greatest number of particles?

1 answer:

3 0

I believe the correct answer would be the last option. All of the quantities given above contain the same number of particles. We determine this by using the avogadro's number. It represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole.

2 moles of carbon atoms ( 6.022 x 10^23 particles / mole ) = 1.20 x10^24 particles

2 moles of carbon dioxide molecules ( 6.022 x 10^23 particles / mole ) = 1.20 x10^24 particles

2 moles of diatomic oxygen molecules ( 6.022 x 10^23 particles / mole ) = 1.20 x10^24 particles

As you can see, no matter what is the gas as long as they have the same number of moles, they would also have same number of particles

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If the molar heat of combustion of liquid benzene at constant volume and 300k is -3272KJ. Calculate the heat of combustion at co

vladimir2022 [97]

Answer:

The heat at constant pressure is -3,275.7413 kJ

Explanation:

The combustion equation is 2C₆H₆ (l) + 15O₂ (g) → 12CO₂ (g) + 6H₂O (l)

$\Delta n_g$ = (12 - 15)/2 = -3/2

We have;

$\Delta H = \Delta U + \Delta n_g\cdot R\cdot T$

Where R and T are constant, and ΔU is given we can write the relationship as follows;

$H = U + \Delta n_g\cdot R\cdot T$

Where;

H = The heat at constant pressure

U = The heat at constant volume = -3,272 kJ

$\Delta n_g$ = The change in the number of gas molecules per mole

R = The universal gas constant = 8.314 J/(mol·K)

T = The temperature = 300 K

Therefore, we get;

H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ

The heat at constant pressure, H = -3,275.7413 kJ.

4 0

2 years ago

Please help me to do this assignment

kirill [66]

Answer:

1. Objective

2. Objective

3. Opinion

4. Objective

5. Opinion

6. Opinion

7. Opinion (I think.)

8. Opinion (I think.)

Explanation:

7 0

3 years ago

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an ex

Kay [80]

The question is incomplete. The complete question is :

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.

Solution :

Consider the equation of the reaction between NaOH and $HCl$

NaOH (aq) + HCl (aq) → NaCl(aq) + $H_2O (l)$

The above equation tells us that $1 \text{mole}$ of $NaOH$ reacts with $1 \text{mole}$ of $HCl$ .

So at the equivalence point, the moles of NaOH added = moles of $HCl$ present.

If the volume of the $HCl$ taken = $V_1$ mL and the conc. of $HCl$ = $M_1$ mole/L

The volume of NaOH added up to the color change = $V_2 \text{ and conc of NaOH = M}_2$ mole/L

Moles of $HCl$ taken = $V_1 \ mL \times M_1 \ mol/100 \ mL = V_2M_2 \times 10^{-3}$ moles.

The color change will occur when the moles of NaOH added is equal to the moles of $HCl$ taken.

Thus when $V_1 M_1 \times 10^{-3} = V_2M_2 \times 10^{-3}$

or when $V_1M_1 = V_2M_2$

or $V_2=\frac{V_1M_1}{M_2}$ mL of NaOH added, we observe the color change.

Where $V_1, M_1$ are the volume and molarity of the $HCl$ taken.

$M_2$ is the molarity of NaOH added.

When both the NaOH and $HCl$ are of the same concentrations, i.e. if $M_1=M_2$ , then $V_2=V_1$

Or the 40 mL of $HCl$ will need 40 mL of NaOH for a color change and

30 mL of $HCl$ would need 30 mL of NaOH for the color change (provided the concentration $M_1=M_2$ )

7 0

3 years ago

37.9 grams of an unknown substance undergoes a temperature increase of

jenyasd209 [6]

Answer:

1.023 J / g °C

Explanation:

m = 37.9 grams

ΔT = 25.0*C

H = 969 J

c = ?

The equation relating these equation is;

H = mcΔT

making c subject of formulae;

c = H / mΔT

c = 969 J / (37.9 g * 25.0*C)

Upon solving;

c = 1.023 J / g °C

4 0

3 years ago

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

6 0

3 years ago