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Sloan [31]
3 years ago
9

A balloon has a volume of 1.75L at a temperature of 298K. What will be the volume of the balloon if you take it out into the win

ter cold air at 258K? Assume pressure is constant.
Chemistry
1 answer:
lozanna [386]3 years ago
7 0

Answer:

1.52 L

Explanation:

P1V1T2=V2P2T1

V2=V1T2/T1

Fill in with given values then solve

We can check this by knowing that V and T at constant P have a proportional relationship. Hence, this is correct.

- Hope that helped! Please let me know if you need further explanation.

You might be interested in
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
The pH of a 0.65M solution of hydrofluoric acid HF is measured to be 1.68. Calculate the acid dissociation constant Ka of hydrof
sergey [27]

Answer:

Kₐ = 6.7 x 10⁻⁴

Explanation:

First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:

HF + H₂O     ⇄   H₃O⁺ +   F⁻

Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]

Since we are given the pH we can calculate the  [ H₃O⁺ ]  ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1  relation , we will also have [F⁻ ]. The  [ HF ] is given in the question so we have all the information that is needed to  compute Kₐ.

pH = -log [ H₃O⁺ ]

1.68 = - log [ H₃O⁺ ]

Taking antilog to both sides of this equation:

10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻²  M= [ H₃O⁺ ]

[ F⁻ ] = 2.1 X 10⁻² M

Solving for Kₐ :

Kₐ = ( 2.1 X 10⁻² ) x  ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴  

(Rounded to two significant figures, the powers of 10 have infinite precision )

4 0
3 years ago
1. The electron configuration below represents the ground state of an atom.
Natalka [10]

Answer:

{S}^{16}

group 16 period 2 of the periodic table

note: that is not the electronic configuration, that is the Bohr model.

4 0
2 years ago
I'll GIVE YOU THE BRAINLIEST!!!!!!!!!!!!!!!!!!!!
ahrayia [7]
Sea arch, headland, sea cave, cliff, sea stack sorry if I’m wrong hope this helped
3 0
3 years ago
Why is the moon unique
Ainat [17]

Answer:

Explanation:

well there is many things in our solar system and a moon is one of them. the moons in our solar system are big and small and orbit our planets. the moon is unique because it shines in the dark while it lets the sun shine in the light.

Thats what i would say but its an eh answer

8 0
3 years ago
Read 2 more answers
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