How many grams are contained in a 0.183 mol sample of ammonium phosphate?

A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a

Andrew [12]

Answer:

Question 1: 1 s after the motion starts

Question 2: 0 (just when the motion starts)

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

1. From the speedometer shown at the right.

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

Free fall equation: Vf = Vo - gt

Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

2. Time when the speedometer displays a reading of 20 m/s

First, calculate the launching speed:

Vf = Vo - gt

Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

0 = Vo - gt

Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

20 m/s = 30 m/s - 10 m/s² × t

t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

3. Time when the speedometer displays a reading of 30 m/s

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

6 0

3 years ago

Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h

andrew11 [14]

Answer : The change in entropy is $6.05\times 10^3J/K$

Explanation :

Formula used :

$\Delta S=\frac{m\times L_v}{T}$

where,

$\Delta S$ = change in entropy = ?

m = mass of water = 1.00 kg

$L_v$ = heat of vaporization of water = $2256\times 10^3J/kg$

T = temperature = $100^oC=273+100=373K$

Now put all the given values in the above formula, we get:

$\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}$

$\Delta S=6048.25J/K=6.05\times 10^3J/K$

Therefore, the change in entropy is $6.05\times 10^3J/K$

5 0

3 years ago

Scientists believe that ancient bacteria represents the first evidence of life on Earth. Which of these best describes the proce

almond37 [142]

It's is reproduction in single called organisms

5 0

3 years ago

Read 2 more answers

When the amount of oxygen is limited, carbon and oxygen react to form carbon monoxide. How many grams of CO can be formed from 3

Alinara [238K]

61.25 grams of CO can be formed from 35 grams of oxygen.

The molecular mass of oxygen is 16 gmol⁻¹

The molecular mass of carbon monoxide is 28 gmol⁻¹

Explanation:

The molar mass of carbon monoxide is molar mass of C added to that of O;

12 + 16 = 28

= 28g/mol

The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol

Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;

35/32 * 2

= 70/32 moles

Then multiply by the molar mass of carbon monoxide;

70/32 * 28

= 61.25 g

4 0

3 years ago

JWL 3.R 1.096776 x 107 m) L) 384.6 nm C) 683.8 nm D) 1282 nm E) >1500 nm Ans: D According to the Rydberg equation, the line w

sesenic [268]

Answer:

91.2 nm

Explanation:

The Rydberg equation is given by the formula

1/ λ = Rh ( 1/ n₁² - 1/ n₂²)

where

λ is the wavelength

Rh is Rydberg constant

and n₁ and n₂ are the energy levels of the transion.

We can see from this equation that the wavelength is inversely proportional to the difference of the squares of the inverse of the quantum numbers n₁ and n₂. It follows then that the smallest wavelength will be given when the the transitions are between the greatest separation between n₁ and n₂ whicg occurs when n1= 1 and n₂= ∞ , that is the greater the separation in energy levels the shorter the wavelength.

Substituting for n₁ and n₂ and solving for λ :

1/λ = 1.0974 x 10⁷ m⁻¹ x ( 1/1² -1/ ∞²) = 1.0974 x 10⁷ m⁻¹ x ( 1/1² - 0) =

λ = 1/1.0974 x 10⁷ m = 9.1 x 10⁻8 m = 91.2 nm

4 0

3 years ago