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2 years ago

The density of krypton gas at 1.21 atm and 50.0°c is ________ g/l.

1 answer:

8 0

1) Ideal gas equation:

pV = nRT

2) Transform the equation to compute density, d = m / V

n = mass in grams / molar mass = m / MM

pV = (m/MM) RT

=> pV = mRT/MM

=> m/V = pMM / (RT)

d = pMM / (RT)

p = 1.21 atm

MM = 83.80 g/mol (this is the atomic mass of krypton element)

T = 50 + 273.15 K = 323.15K

3) Compute:

d = (1.25 atm * 83.80 g/mol) / (0.0821 atm*liter /K*mol * 323.15K)

d = 3.95 g / liter

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When 0.270 mol of a nondissociating solute is dissolved in 410.0 mL of CS2, the solution boils at 47.52 ∘C. What is the molal bo

grandymaker [24]

Answer:

Kb = 0.428 m/°C

Explanation:

To solve this problem we need to use the <em>boiling-point elevation formula</em>:

<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m

Where <em>Tsolution</em> and <em>Tpure solvent</em> are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.

So in order to use that equation and solve for Kb, first we <em>calculate the molality of the solution</em>.

molality = mol solute / kg solvent

Density of CS₂ = 1.26 g/cm³

Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg

molality = 0.270 mol / 0.5166 kg = 0.5226 m

Now we <u>solve for Kb</u>:

<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m

47.52 °C - 46.3 °C = Kb * 0.5226 m

Kb = 0.428 m/°C

3 0

2 years ago

Find the number of moles of water that can be formed if you have 170 mol of hydrogen gas and 80 mol of oxygen gas. Express your

Sav [38]

<u>Answer:</u> The amount of water that can be formed is 160 moles

<u>Explanation:</u>

We are given:

Moles of hydrogen gas = 170 moles

Moles of oxygen gas = 80 moles

The chemical equation for the reaction of hydrogen gas and oxygen gas follows:

$2H_2+O_2\rightarrow 2H_2O$

By Stoichiometry of the reaction:

1 mole of oxygen gas reacts with 2 moles of hydrogen gas

So, 80 moles of oxygen gas will react with = $\frac{2}{1}\times 80=160mol$ of hydrogen gas

As, given amount of hydrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of oxygen gas produces 2 moles of water

So, 80 moles of oxygen gas will produce = $\frac{2}{1}\times 80=160moles$ of water

Hence, the amount of water that can be formed is 160 moles

3 0

2 years ago

Why are the concentrations of [H3O+] and [OH−] equal in pure water?

kotegsom [21]

Answer : The correct option is, (A) $[H_3O^+]=[OH^-]$ because one of each is produced every time an $[H^+]$ transfers from one water molecule to another.

Explanation :

As we know that, when the two water molecule combine to produced hydronium ion and hydroxide ion.

The balanced reaction will be:

$HOH+HOH\rightarrow H_3O^++OH^-$

Acid : It is a substance that donates hydrogen ion when dissolved in water.

Base : It is a substance that accepts hydrogen ion when dissolved in water.

From this we conclude that, the hydrogen ion are transferred from one water molecule to the another water molecule to form hydronium ion and hydroxide ion. In this reaction, one water molecule will act as a base and another water molecule will act as an acid.

Hence, the correct option is, (A)

3 0

2 years ago

Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli

Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>

3 0

2 years ago

Consider the reaction. X ( g ) + Y ( g ) − ⇀ ↽ − Z ( g ) K p = 1.00 at 300 K In which direction will the net reaction proceed fo

marta [7]

Answer:

Explanation:

We have in this question the equilibrium

X ( g ) + Y ( g ) ⇆ Z ( g )

With the equilibrium contant Kp = pZ/(pX x pY)

The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional

pV = nRT ⇒ p = nRT/V and n/V is molarity.

Therefore we can calculate the reaction quotient Q

Qp = pZ/(pX x pY) = 1/ 1 x 0.5 atm = 2

Since Qp is greater than Kp the system proceeds from right to left.

We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.

4 0

2 years ago