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Svetlanka [38]
3 years ago
7

What are the three components of the fire triangle?

Chemistry
1 answer:
9966 [12]3 years ago
5 0

Answer:

oxygen, heat and fuel

Explanation:

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The microscopes used today are just like the ones used by Leeuwenhoek and Hooke. True or false
Dmitry [639]

Answer:

False

Explanation:

While we do know that A. Leeuwenhoek used a simple microscope that consisted of only 1 lens, Hooke used a compound microscope. Although, after trying a compound microscope, Hooke found out that it strained his eyes and continued to use a simple microscope for his <em>Micrographia</em>.

Thus, we can say that the (compound) microscopes used today are different than the (simple) microscope used by Hooke and Leeuwenhoek.

6 0
2 years ago
DNA replication is continuous on the (blank)
ExtremeBDS [4]

I don't know anything sorry for inconvenience but I will be used to make a system running with artificial intelligence

4 0
3 years ago
How many molecules are there in 6.73 moles of phosphorus trichloride? (write your answer in scientific notation).
anyanavicka [17]

Answer: There are 4.05\times 10^{24}  molecules in 6.73 moles of phosphorus trichloride

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of PCl_3 contains = 6.023\times 10^{23}  molecules

Thus 6.73 moles of PCl_3 contains = \frac{6.023\times 10^{23}}{1}\times 6.73=4.05\times 10^{24}  molecules

Thus there are 4.05\times 10^{24}  molecules in 6.73 moles of phosphorus trichloride

3 0
3 years ago
A compound of carbon and hydrogen having one double covalent bond between two of the carbon atoms in the molecule is known as wh
Alexandra [31]
It would be known as a meth_(insert - ane, - ene, and -yne) hydrocarbon. 1 bond = -ane 2 bonds = - ene And so on.
7 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
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