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Ludmilka [50]
3 years ago
7

A 250-mL glass bottle was filled with 242mL of water at 20°C and tightly capped. it was then left outdoor overnight, where the a

verage temperature was -5°C. predict what would happen. the density of water at 20°C is 0.998g/cm^3 and that of ice at -5°C is 0.916g/cm^3.
Chemistry
2 answers:
garri49 [273]3 years ago
3 0
It would freeze. yw btw
Softa [21]3 years ago
3 0

Answer:

The glass bottle will get crack.

Explanation:

Volume of the glass bottles,V = 250 mL

Density of water at 20°C  = 0.998g/cm^3

Mass of water in bottle = M

Volume of water at 20°C ,V' = 250 mL (1cm^3= 1 mL)

V>V'

Density=0.988 g/cm^3=\frac{M}{242 cm^3}

M = 239.096 g

Since mass remains constant, the mass of liquid water will be equal to mass of water when freezes

Density of ice at -5°C = 0.916g/cm^3

Volume of the ice at  -5°C= V''

Density=0.916 g/cm^3=\frac{239.096g}{V'}

V''=261.0218cm^3=261.0218 mL

V''>V>V'

As we can see that volume of water has expanded on freezing and exceeded the volume of the glass bottle. This expansion will result in breaking or cracking of a glass bottle.

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2 Carbon double bonds
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What is the Bronsted acid of H2PO4- + OH-. ---- HPO42- + H2O?​
RoseWind [281]

Answer:

The bronsted- Lowry acid is H₂PO₄⁻

Explanation:

Bronsted-Lowry acid  donates a proton (H⁺)

H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O

In the reaction above, H₂PO₄⁻ is donating the proton to OH⁻ resulting in H₂O and the deprotonated species. This makes it a bronsted-Lowry acid.

7 0
3 years ago
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For elements in the third row of the periodic table and beyond, the octet rule is often not obeyed. a friend of yours says this
Rom4ik [11]

The reason that some of the elements of period three and beyond are steady in spite of not sticking to the octet rule is due to the fact of possessing the tendency of forming large size, and a tendency of making more than four bonds. For example, sulfur, it belongs to period 3 and is big enough to hold six fluorine atoms as can be seen in the molecule SF₆, while the second period of an element like nitrogen may not be big to comprise 6 fluorine atoms.  

The existence of unoccupied d orbitals are accessible for bonding for period 3 elements and beyond, the size plays a prime function than the tendency to produce more bonds. Hence, the suggestion of the second friend is correct.  


4 0
2 years ago
A mixture of three noble gases has a total pressure of 1.25 atm. The individual pressures exerted by neon and argon are 0.68
FinnZ [79.3K]

First of all, as you seen the gases are noble which means that will not react with each other and in this case each gas create individual pressure.  

P_{T}= total pressure  

P_{Ne} = pressure of neon  

P_{Ar} = pressure of argon  

P_{He} = pressure of helium {which is required}

P_{T} = P_{Ne} + P_{Ar} + P_{He}    

1.25 = 0.68 + 0.35 +  P_{He}

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4 0
3 years ago
3.
sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5.  V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a)  ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0
3 years ago
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