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3 years ago

How many moles are there in 3.612 x 1024 formula units of Na2SO4?

Chemistry

1 answer:

vitfil [10]3 years ago

5 0

Answer:

<h3>Theanswer is 6 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.612 \times {10}^{24} }{6.02 \times {10}^{23} } \\$

We have the final answer as

<h3>6 moles</h3>

Hope this helps you

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A 0.5 kg ball rolls down a frictionless inclined plane. The height of the inclined plane is 0.5m. What is the kinetic energy of

joja [24]

Answer:

2.45Joules

Explanation:

since kinetic energy is K.E.=1/2m $v^{2}$

lost in potential energy is gain of kinetic energy

mgh=1/2mv^2

9.8*2*0.5=v^2

$v^{2}$ =9.8

v= $\sqrt{9.8}$ =3.13m/s

now k.e. = 0.5*m* $v^{2}$ =0.5*0.5*9.8 =2.45J

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3 years ago

Which of the following is the correct name for the compound HF?

alexira [117]

Name for the compound HF is Hydrogen fluoride.
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3 years ago

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A solution of ph 5 has _____ times more hydrogen ions than that of pure water.

Alik [6]

PH measures the concentration of hydrogen ions on the log 10 scale. Thus, a pH 5 solution has 2 order of magnitude difference from that of pure water, which has a pH of 7.

Therefore, $10^{2} = 100$ . A solution of pH 5 has 100 times more hydrogen ions that that of pure water.

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3 years ago

How to round significant figures

marissa [1.9K]

Explanation:

To round to a significant figure:

look at the first non-zero digit if rounding to one significant figure
look at the digit after the first non-zero digit if rounding to two significant figures
draw a vertical line after the place value digit that is required
look at the next digit
if it's 5 or more, increase the previous digit by one
if it's 4 or less, keep the previous digit the same
fill any spaces to the right of the line with zeros, stopping at the decimal point if there is one

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3 years ago

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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