How many moles can be found in 1.5 x 10^24 atoms of Calcium?

fills a 500.mL flask with 3.6atm of carbon monoxide gas and 1.2atm of water vapor. When the mixture has come to equilibrium she

enot [183]

Answer:

The answer to the question is

The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits

Explanation:

To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure

At the first trial the mixture contains

3.6 atm CO

1.2 atm H₂O (g)

Total pressure = 3.6+1.2= 4.8 atm

which gives

3.36 atm CO

0.96 atm H₂O (g)

0.24 atm H₂ (g)

That is

CO+H₂O→CO(g)+H₂ (g)

therefore the mixture contained

0.24 atm CO₂ and the total pressure =

3.36+0.96+0.24+0.24 = 4.8 atm

when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂

At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857

adding 1.8 atm CO gives 4.46 atm hence we have

(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857

which gives x = 0.031 atm or x = -0.6183 atm

Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm

7 0

3 years ago

A particular sample of pure iron requires 0.612 kJ of energy to raise its temperature from 30.°C to 51°C. What must be the mass

mixas84 [53]

Answer:

m = 65.637 g

Explanation:

Heat = 0.612 kJ = 612 J ( Converting to J by multiplying by 1000)

Initial Temperature = 30.°C

Final Temperature = 51°C

Temperature change = Final Temperature - Initial Temperature = 51 - 30 = 21°C

Mass = ?

The relationship between these quantities is given by the equation;

H = mCΔT

where c = 0.444 J/g°C

Inserting the values in the equation;

612 = m * 0.444 * 21

m = 612 / (0.444 * 21)

m = 65.637 g

3 0

3 years ago

A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

kaheart [24]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Explanation:

Mass of an alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Mass of water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

$m_{a}$ $C_{a}$ [ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This is the specific heat of the alloy.

6 0

3 years ago

Sodium hydroxide (NaOH) and hydrochloric acid (HCl) combine to make table salt (NaCl) and water (H2O). Which equation is balance

Art [367]

The equation that is balanced among the given options is the first equation NaOH + HCl → NaCl + H2O

From the question,

We are to determine which of the options gives a balanced equation for the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) to make table salt (NaCl) and water (H₂O)

To do this, we will check which of the given equations is balanced, by checking the equations one after the other.

First, we will define what is meant by balanced equation.

Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products

For NaOH + HCl → NaCl + H2O

On the reactants side

Na = 1

O = 1

H = 2

Cl = 1

On the products side

Na = 1

O = 1

H = 2

Cl = 1

Since equal number of atoms of each element are present at both the reactants and products sides, then the equation is balanced

For 2NaOH + 2HCl → 2NaCl + H2O

On the reactants side

Na = 2

O = 2

H = 4

Cl = 1

On the products side

Na = 2

O = 1

H = 2

Cl = 2

The number of atoms of each element present at the reactants side and that present at products side are not equal.

∴ The equation is not balanced

For 2NaOH + HCl → NaCl + H2O

On the reactants side

Na = 2

O = 2

H = 3

Cl = 1

On the products side

Na = 1

O = 1

H = 2

Cl = 1

The number of atoms of each element present at the reactants side and that present at products side are not equal.

∴ The equation is not balanced

For NaOH + 2HCl → NaCl + H2O

On the reactants side

Na = 1

O = 1

H = 2

Cl = 2

On the products side

Na = 1

O = 1

H = 2

Cl = 1

The number of atoms of each element present at the reactants side and that present at products side are not equal.

∴ The equation is not balanced

Hence, the equation that is balanced is the first equation NaOH + HCl → NaCl + H2O

Learn more here: brainly.com/question/17015942

6 0

2 years ago

Read 2 more answers

What percent by mass of aluminum is present in aluminum sulfate ?

xxMikexx [17]

The percent of aluminium of a aluminium sulfate, AI2(SO4)3 is 15.8%.

4 0

3 years ago