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Vlad1618 [11]
3 years ago
7

Carbon tetrachloride is an iconic compound true or false

Chemistry
2 answers:
zmey [24]3 years ago
7 0
The answer is false :))
il63 [147K]3 years ago
4 0

Answer:

False

Explanation:

Carbon tetrachloride is a covalent compound and not an ionic compound.

This is because it has four nonpolar covalent bonds between carbon and chlorine.

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If two gases with pressures of 2 atm and 3 atm are mixed at a constant temperature, what will the total pressure be?
slamgirl [31]
If two gases with pressures of 2 atm and 3 atm are mixed at constant temperature, the total pressure will be the sum of the two pressures. Therefore the answer is D. 2 atm + 3 atm or 5 atm will be the total pressure of the gas mixture. 
8 0
3 years ago
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20. Given the reaction at equilibrium:
Gemiola [76]
PV = nRT

If pressure increases, so will moles.

The answer is 2) Increase.

8 0
2 years ago
The volume of a sample of gas at 273 K is 200. liters. If the volume is doubled at constant pressure, what will be the new tempe
gayaneshka [121]

Answer:

New temperature = 546 K

Explanation:

Given that,

Initial volume, V = 200 L

Initial temperature, T = 273 K

We need to find the new temperature if the volume is doubled.

The relation between temperature and volume is direction. If volume is doubled, it means the temperature gets also doubled.

New temperature of the gas = 2(273) = 546 K

8 0
3 years ago
An air stream enters a variable area channel at a velocity of 30 m/s with a pressure of 120 kPa and a temperature of 10°C. At a
nekit [7.7K]

Answer:

P₂= 74 kPa under constant density and ρ₂ = 1.06 kg/m³ (-38.6% of error compared with incompresible assumption) . Thus Bernoulli’s equation should not be applied

Explanation:

Assuming ideal gas behaviour of air , then

P*V= n*R*T = m / M * R *T

since

ρ= m/V = P*M /( R *T)

where

n= moles , V= volume , m= mass

ρ= density

P= pressure = 120 kPa= 120000 Pa

M= molecular weight of air = 0.21*32+0.79*28= 28.24 gr/mol = 0.02824 kg/mol

T= absolute temperature = 10°C + 273 = 283 K

R= ideal gas constant = 8.314 J/mol K

solving for ρ

ρ=  P*M /( R *T) = 120000 Pa*0.02824 kg/mol/(8.314 J/mol K*283 K) = 1.47 kg/m³

then from Bernoulli's equation

P₁ + ρ*v₁²/2 = P₂ + ρ*v₂²/2

where 1 denotes inlet and 2 denotes other point , p = pressure and v= velocity . Then solving for p₂

P₁ + ρ*v₁²/2 = P₂ +ρ*P₂²/2

P₂=  P₁ +ρ*v₁²/2 - ρ*v₂²/2  = P₁ +ρ/2*(v₁² - v₂²)

replacing values

P₂= P₁ +ρ/2*(v₁² - v₂²) = 120000 Pa + 1.47 kg/m³/2*[(30 m/s)²-(250 m/s)²] = 74724 Pa = 74 kPa

P₂= 74 kPa

then if the temperature remains constant

ρ₁= P₁*M /( R *T) and ρ₂= P₂*M /( R *T)

dividing both equations

ρ₂/ρ₁ = P₂/ P₁

ρ₂ = (P₂/ P₁)*ρ₁

then from Bernoulli's equation

P₁ + ρ₁*v₁²/2 = P₂ + ρ₂*v₂²/2

P₂ = P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2

therefore

ρ₂ = (P₂/ P₁)*ρ₁ = (P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2 ) /P₁ *ρ₁

P₁ * ρ₂  = P₁ *ρ₁  + ρ₁²*v₁²/2 - ρ₂*ρ₁ * v₂²/2

P₁ * ρ₂ + ρ₂*ρ₁ * v₂²/2  = P₁ *ρ₁  + ρ₁²*v₁²/2

ρ₂* (P₁ + ρ₁ * v₂²/2) = P₁ *ρ₁  + ρ₁²*v₁²/2

ρ₂ = (P₁ *ρ₁  + ρ₁²*v₁²/2)/(P₁ + ρ₁ * v₂²/2) =  (P₁ + ρ₁*v₁²/2)/(P₁/ρ₁ + v₂²/2)

replacing values

ρ₂ = ( 120000 Pa + 1.47 kg/m³/2*(30 m/s)²)/(120000 Pa/1.47 kg/m³+1/2*(250 m/s)²)

ρ₂ = 1.06 kg/m³

the error of assuming constant ρ would be

e = (ρ₂ - ρ)/ρ₂=  1- ρ/ρ₂= 1- 1.47 kg/m³/1.06 kg/m³ = -0.386 (-38.6%)

thus Bernoulli’s equation should not be applied

8 0
3 years ago
mercury metal is poured into a graduated cylinder that holds exactly 22.5 mL. the mercury used to fill the cylinder weighs 306.0
SpyIntel [72]
306.0g/22.5mL= 13.6 mL/g
D=m/v
The mass is 306.0
The volume is 13.6
4 0
3 years ago
Read 2 more answers
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