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3 years ago

14

A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains 1.50 g of carbon and 2.00 g of oxygen. Anoth

er sample of ascorbic acid isolated from citrus fruits contains 6.25 g of carbon.?

1 answer:

Karo-lina-s [1.5K]3 years ago

6 0

Answer:

8.467 gm

Explanation:

The law governing this problem is "The Law of Constant Composition "

As per this law, all compounds irrespective of their origin and source have the same composition and properties in their purest form

It is a simple proportion and ratio related problem.

1.50 grams of carbon require 2.0 grams of oxygen

1.0 grams of carbon will require oxygen

= $\frac{2}{1.5}$

6.35 grams of carbon will require oxygen

$\frac{2}{1.5}\times6.35$ = 8.4666\\= 8.467

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A roller coaster car rapidly picks up speed as it rolls down a slope as it starts down the slope its speed is 4m/s but 3 seconds

Gwar [14]

Answer:

The acceleration is 6 [m/s^2]

Explanation:

We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.

$v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\$

Now replacing the values we have:

$a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]$

3 0

2 years ago

Calculate the change in the energy of an electron that moves from the n = 3 level to the n = 2 level. What type of light is emit

marissa [1.9K]

Answer:

Red light

Explanation:

The energy emitted during an electron transition in an atom of hydrogen is given by

$E=E_0 (\frac{1}{n_2^2}-\frac{1}{n_1^2})$

where

$E_0 = 13.6 eV$ is the energy of the lowest level

n1 and n2 are the numbers corresponding to the two levels

Here we have

n1 = 3

n2 = 2

So the energy of the emitted photon is

$E=(13.6) (\frac{1}{2^2}-\frac{1}{3^2})=1.9 eV$

Converting into Joules,

$E=(1.9 eV)(1.6\cdot 10^{-19} J/eV)=3.0\cdot 10^{-19} J$

And now we can find the wavelength of the emitted photon by using the equation

$E=\frac{hc}{\lambda}$

where h is the Planck constant and c is the speed of light. Solving for $\lambda$ ,

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.0\cdot 10^{-19}}=6.63\cdot 10^{-7} m = 663 nm$

And this wavelength corresponds to red light.

5 0

2 years ago

A firecracker in a coconut blows the coconut into three pieces. Twopieces of equal mass fly off south and west, perpendicular to

loris [4]

Answer:

V = 13m/s

North-East (i.e. 45 degree from North)

Explanation:

This question deals with the idea of momentum.

Since the directions of a compass are fixed, we can take south as horizontal and west as vertical. Since both pieces of coconut are of equal mass, then the resultant of the two pieces would be in between South and West and the third piece would be opposite this direction and be in the North-East (i.e. 45 degree from North) direction

To find the speed of the third piece, we first find the speed on the two pieces of the coconut in terms of $V_{x}$ (horizontal component of velocity) and $V_{y}$ (vertical component of velocity)

We have

$V_{y}$ = Velocity in South direction = 18m/s

$m_{y}$ = Mass of piece in South direction = m

$V_{x}$ = Velocity in West direction = 18m/s

$m_{x}$ = Mass of piece in West direction = m

$V_{ry}$ = Velocity of Third Piece in North direction = unknown

$V_{rx}$ = Velocity of Third Piece in East direction = unknown

$m$ = Mass of third piece = 2m

$mV_{ry} = m_{y} V_{y} \\ 2mV_{ry} = m18\\ V_{ry} = 9$

$mV_{rx} = m_{x} V_{x} \\ 2mV_{rx} = m18\\ V_{rx} = 9$

Since we now know the horizontal and vertical component of the velocity of the third piece of coconut we find the resultant velocity. Since we know the direction is North-East, we can imagine a right angled triangle with base of 9 and height 9 and the hypotenuse equal to the resultant velocity .

We can simply apply the Pythagoras theorem and find the hypotenuse

$Hypotenuse^{2} = Height^{2} + Base^{2} \\ V_{3}^2 = 9^{2} + 9^{2} \\ V = 12.728 m/s$

V = 13m/s

6 0

3 years ago

Quick please and will give Brainliest!!!

masha68 [24]

25 nC

That is the answer

3 0

3 years ago

Read 2 more answers

zysi [14]

<h3>No:1</h3>

The object is moving with constant or uniform acceleration and in average speed

<h3>No:-2</h3>

The object is de accelerating

<h3>No:-3</h3>

The object deaccelerated and came to rest so fast.

<h3>No:-4</h3>

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<h3>No:-5</h3>

The object accelerated at first so fast then move with constant acceleration then again accelerated .

7 0

2 years ago