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1 year ago

Why is a solution of 4% acetic acid in 95% ethanol used to wash the crude aldol-dehydration product?

1 answer:

5 0

A cold acetic acid solution is used to wash the residue of the reagent in preparation of an aldol condensation product after vacuum filtration. The main reason in washing with the acetic acid rinse is to neutralize any sodium hydroxide.

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In which case does viscosity play a dominant role? Case A: a typical bacterium (size ~ 1 mm1 mm and velocity ~ 20 mm/s20 mm/s) i

My name is Ann [436]

Answer:

Case A

Explanation:

given,

size of bacteria = 1 mm x 1 mm

velocity = 20 mm/s

size of the swimmer = 1.5 m x 1.5 m

velocity of swimmer = 3 m/s

Viscous force

$F = \eta A \dfrac{dv}{dx}$

for the bacteria

$F = \eta \times 10^{-6}\times 20\times 10^{-3}$

$F =2\times 10^{-8} \eta\ N$

for the swimmer

$F = \eta \times 1.5^2\times 3$

$F =6.75 \eta\ N$

from the above force calculation

In case B inertial force that represent mass is more than the inertial force in case of bacteria.

Viscous force is dominant in case of bacteria.

So, In Case A viscous force will be dominant.

5 0

1 year ago

Which type of listening response includes the use of head nods, facial expressions, and short utterances such as "uh-huh" that s

BARSIC [14]

This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others.

4 0

1 year ago

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

$Mm=44.0 g/mol$ is the molar mass of the carbon dioxide

So, the number of moles of the gas is

$n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol$

Now we can find the pressure of the gas by using the ideal gas equation:

$pV=nRT$

where

p is the pressure

$V=1.00 L = 0.001 m^3$ is the volume

n = 0.023 mol is the number of moles

$R=8.314 J/mol K$ is the gas constant

$T=25.0^{\circ}+273=298 K$ is the temperature of the gas

Solving the equation for p, we find

$p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa$

And since we have

$1 atm = 1.01\cdot 10^5 Pa$

the pressure in atmospheres is

$p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm$

5 0

2 years ago

A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizonta

shusha [124]

Answer:

19.99 kg m²/s

Explanation:

Angular Momentum (L) is defined as the product of the moment of Inertia (I) and angular velocity (w)

L = m r × v.

r and v are perpendicular to each other,

where r = lsinθ.

l = 2.4 m

θ= 34°

g = 9.8 m/s² and m = 5 kg

resolving using newtons second law in the vertical and horizontal components.

T cos θ − m g = 0

T sin θ − mw² lsin θ = 0

where T is the force with which the wire acts on the bob

w = √g / lcosθ

= √ 9.8 / 2.4 ×cos 34

= 2.2193 rad/s

the angular momentum L = mr× v

= mw (lsin θ)²

= 5 × 2.2193 (2.4 ×sin 34°)²

=19.99 kg m²/s

8 0

2 years ago

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

adell [148]

Answer:

$F_{x}=2.31N$ to the right.

$F_{y}=2.1N$ to in the upwards direction.

Explanation:

In order to solve this problem, we must first start by drawing a diagram of the situation. (See attached diagram).

So, remember that a force is determined by multiplying the mass of the parcticle by its acceleration:

F=ma

so in order to find the components of the force, we need to start by finding its acceleration.

Acceleration is found by using the following formula:

$a=\frac{V_{f}-V{0}}{t}$

so we can subtract the two vectors, like this:

$a=\frac{(6.00i+4.0j)m/s-1.60i m/s}{8s}$

which yields:

$a=\frac{(4.4i+4.0j)m/s}{8s}$

or:

$a=(0.55i + 0.5j) m/s^{2}$

so now I can find the components of the force:

$F=(4.2kg)(0.55i + 0.5j) m/s^{2}$

which yields:

F=(2.31i+2.1j)N

so the components of the force are:

$F_{x}=2.31N$ to the right.

$F_{y}=2.1N$ to in the upwards direction.

6 0

1 year ago