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konstantin123 [22]
2 years ago
5

How many molecules are in 502.1 g of NaF?

Chemistry
1 answer:
Murrr4er [49]2 years ago
7 0

Answer:

11.958129200057586 molecules

Explanation:

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A certain object has a volume of 25.0 ml and a mass of 100 g. what is the density of the object
Tpy6a [65]
Density= mass/volume

             = 100/25
density  = 4g/ml
8 0
3 years ago
How many moles of oxygen are needed to react with 87 grams of aluminum
labwork [276]

Answer:

2.4 moles of oxygen are needed to react with 87 g of aluminium.

Explanation:

Chemical equation:

4Al(s)  + 3O₂(l)   → 2AlO₃(s)

Given data:

Mass of aluminium = 87 g

Moles of oxygen needed = ?

Solution:

Moles of aluminium:

Number of moles of aluminium= Mass/ molar mass

Number of moles of aluminium= 87 g/ 27 g/mol

Number of moles of aluminium= 3.2 mol

Now we will compare the moles of aluminium with oxygen.

                              Al         :         O₂

                               4          :         3

                               3.2       :         3/4×3.2 = 2.4 mol

2.4 moles of oxygen are needed to react with 87 g of aluminium.

5 0
2 years ago
T/F___ At the eutectic composition, an alloy can solidify at a constant temperature.___ For effective dispersion strengthening,
azamat

Answer:

  • TRUE
  • FALSE
  • TRUE
  • FALSE
  • FALSE
  • TRUE

Explanation:

  • At the eutectic composition, an alloy can solidify at a constant temperature : TRUE . this is because at eutectic composition the type of reaction that takes place there is invariant reaction in its thermal equilibrium
  • For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round : FALSE. because the rounded shape will not cause a crack.
  • Intermetallic compounds are usually hard and brittle : TRUE. because Intermetallic compounds prevents dislocation movements and this makes them brittle and hard
  • For the effective dispersion and strengthening, the dispersed phase should be continuous : FALSE. this is because the dispersed precipitate must be small and not continuous
  • Stoichiometric intermetallic compounds exist over a range of compositions : FALSE
  • Faster solidification results in smaller interlamellar spacing : TRUE
4 0
2 years ago
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: <u>1 s after the motion starts</u>

Question 2: <u>0 (just when the motion starts)</u>

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.

<u>1. From the speedometer shown at the right.</u>

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.

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

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.

<u />

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

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
Is the salt solution in the Dead<br>Sea dilute, concentrated or<br>saturated? Explain.​
dexar [7]
The waters of the Dead Sea are extremely saline, and, generally, the concentration of salt increases toward the lake's bottom. ... The deep water was saturated with sodium chloride , which precipitated to the bottom.
5 0
2 years ago
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