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3 years ago

What is the density of hydrogen sulfide at 1.1 atm and 304K?

Chemistry

1 answer:

Daniel [21]3 years ago

6 0

Use equation:

PV = nRT

1.4*V = 1*0.082057*332

V = 27.24/1.4

V = 19.46L

Molar mass H2S = 2+32 = 34g/mol

34g H2S has volume 19.46L

Density = mass / volume = 34/19.46 = 1.75 g/L

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How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride (FM 121.135) to give a pH of 7.60 in a fina

liubo4ka [24]

Answer:

41.64mL of NaOH 0.500M must be added to obtain the desire pH

Explanation:

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.

Replacing in H-H equation:

7.60 = 8.072 + log [A⁻] / [HA]

0.3373 = [A⁻] / [HA] (1)

10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:

10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:

[A⁻] + [HA] = 0.08255 (2)

Replacing (1) in (2):

0.3373 = 0.08255 - [HA] / [HA]

0.3373[HA] = 0.08255 - [HA]

1.3373[HA] = 0.08255

[HA] = 0.06173 moles

Thus:

[A⁻] = 0.08255 - 0.06173 = 0.02082 moles [A⁻]

The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:

HA + NaOH → A⁻ + H₂O + K⁺

Thus, you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. As NaOH solution is 0.500M:

0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =

<h3>41.64mL of NaOH 0.500M must be added to obtain the desire pH</h3>

3 0

3 years ago

Your science teacher gives you three liquids to pour into a jar. After pouring all of them into the jar, the liquids layer as se

sweet-ann [11.9K]

Answer:

density

Explanation:

6 0

3 years ago

the determination of a chemical formula n this experiment, you will use the law of definite proportions to find the chemical for

victus00 [196]

Answer: Strictly a laboratory analysis and can only be done using the data obtained during analysis

Explanation:

To find a solution to this problem, you need to use the data collected during the lab work. A guide could be finding the possible forms of hydrated copper chlorides in reference books. Since it's also a lab work, you can definitely compare your data with lab mates.

The formula CuxCly.zH₂O and its name chloride hydrate already gives you an idea of the possibilities of the value of the integers, hence you can take a good guess for the identity of the unknown salt and calculate the theoretical formular weight for it. From the that you can proceed to also find the mass of water and copper from your lab analysis.

7 0

3 years ago

At 400 K oxalic acid decomposes according to the reaction:H2C2O4(g)→CO2(g)+HCOOH(g)In three separate experiments, the intial pre

padilas [110]

Answer:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

Explanation:

For the reaction:

H₂C₂O₄(g) → CO₂(g) + HCOOH(g)

At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:

H₂C₂O₄(g) = P₀ - x

CO₂(g) = x

HCOOH(g) = x

P at t=20000 is:

P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x

For 1st point:

x = 92,8-65,8 = 27

Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8

2nd point:

x = 130-92,1 = 37,9

H₂C₂O₄(g): 92,1 - 37,9 = 54,2

3rd point:

x = 157-111 = 46

H₂C₂O₄(g): 111-46 = 65

Now, as the rate law is :

v = k P[H₂C₂O₄]

Based on integrated rate law, k is:

(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k

1st point:

k = 2,64x10⁻⁵

2nd point:

k = 2,65x10⁻⁵

3rd point:

k = 2,68x10⁻⁵

The averrage of this values is:

k = 2,66x10⁻⁵

That means law is:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

I hope it helps!

4 0

3 years ago

What are the processes for the 3 types of weathering

sweet [91]

The processes for the 3 types of weathering are,
biological: rocks worn down by plants
physical: water freezing and thawing in cracks of rocks
chemical: Acid rain
Hope this helped!!

7 0

3 years ago