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ivanzaharov [21]
3 years ago
7

As part of a chemistry experiment, Austin ends up with a beaker containing a mixture of two liquids, which he knows have boiling

points of 77 °C and 103 °C. Austin is already familiar with___________ as a good method for separating liquids, so he asks his chemistry teacher where to find the heating mantle and condensing tube he'll need to___________ the mixture.
Engineering
1 answer:
asambeis [7]3 years ago
7 0

Answer:

Distillation, heat

Explanation:

Here in this question, we simply want to look at the best options that could fit in the gaps.

We have a mixture of liquids having boiling points which is far from each other.

Whenever we have a mixture of liquids with boiling points far away from each other, the best technique to use in separating them is to use distillation. That is why we have that as the best fit for the first missing gap.

Now, to get the liquids to separate from each other, we shall be needing the heating mantle for the application of heat. This ensures that the mixture is vaporized. After vaporization, the condensing tube will help to condense the vapor of each of the liquids once we reach the boiling point of either of the two.

Kindly note that the liquid with the lower temperature will evaporate first and will be first obtained. In fact after reaching a little above the boiling point of the lower boiling liquid, we can be sure that what we have left in the mixture pot is the second other liquid with the higher boiling point.

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Semmy [17]

Answer:

I=9.6×e^{-8}  A

Explanation:

The magnetic field inside the solenoid.

B=I*500*muy0/0.3=2.1×e ^-3×I.

so the total flux go through the square loop.

B×π×r^2=I×2.1×e^-3π×0.025^2

=4.11×e^-6×I

we have that

(flux)'= -U

so differentiating flux we get

so the inducted emf in the loop.

U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)

so, I=2.9×e^{-6}÷30

I=9.6×e^{-8}  A

6 0
3 years ago
Establishes general guidelines concerning licensing and vehicle
gavmur [86]

Answer:

A. National Highway Safety Act

Explanation:

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5 0
3 years ago
Read 2 more answers
Which of the following is NOT associated with Urban Sprawl?
pochemuha

The option that is not associated with the given term called urban sprawl is; Option A: Blocking high views

What is Urban Sprawl?

Urban sprawl is defined as the rapid expansion of the geographic boundaries of towns and cities which is often accompanied by low-density residential housing and increased reliance on the private automobilefor movement.

Looking at the given options, "blocking high views" is the option that is not typically a problem associated with urban sprawl because urbanization usually takes place on relatively flat levels.

The missing options are;

a. blocking high views

b. destroying animal habitats

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8 0
2 years ago
Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin
Vladimir79 [104]

Answer:

a)

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

b)

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

c)

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

a)

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 =  1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 =  1.013 μm

b)

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 =  0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.243 μm

c)

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 =  0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.608 μm

6 0
3 years ago
Integer to Float Conversion All labs must be done during lab time. Each labs worth 10 points The lab can be hand in next day wit
andrew-mc [135]

Answer:

Code explained below

Explanation:

.data

msg1: .asciiz "Please input a temperature in celsius: "

msg2: .asciiz "The temperature in Fahrenheit is: => "

num: .float 0.0

.text

main:

#print the msg1

li $v0, 4

la $a0, msg1

syscall

#read the float value from user

li $v0,6 #read float syscall value is $v0

syscall #read value stored in $f0

#formula for celsius to fahrenheit is

#(temperature(C)* 9/5)+32

#li.s means load immediate float

#copy value 9.0 to $f2

li.s $f2,9.0  

#copy value 5.0 to $f3

li.s $f3,5.0

# following instructions performs: 9/5

#div.s - division of two float numbers

#divide $f2 and f3.Result will stores in $f1

div.s $f1,$f2,$f3

#following instruction performs: temperature(C) * (9/5)

#multiple $f1 and $f0.Result stored in $f1

mul.s $f1,$f1,$f0

#copy value 32 to $f4

li.s $f4,32.0

#following instruction performs: (temperature(C) * (9/5))+32

#add $f1 and $f4.Result stores in $f1

add.s $f1,$f1,$f4

#store float from $f1 to num

s.s $f1,num

#print the msg2

li $v0, 4 #print string syscall value is 4

la $a0, msg2 #copy address of msg2 to $a0

#print the float

syscall

li $v0,2 #print float syscall value is 2

l.s $f12,num #load value in num to $f12

syscall

#terminate the program

li $v0, 10 #terminate the program syscall value is 10

syscall

4 0
3 years ago
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