When looking across the period table from left to right, which is the first group that contains nonmetals?

Locate the mode of 12, 3, 5, 17, 3, 18, 5, 11, 11, 15, 3, 9, 3.

zimovet [89]

Answer:
Mode = 3 because it is listed 4 times

8 0

3 years ago

If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat

Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

429 g = 429 mL

429 / 1000 = 0.429 L of water

3 0

3 years ago

The diagram below shows the stars that are nearest to our solar system.

abruzzese [7]

Answer:

no u tried of the same dam

thing

Explanation:

7 0

3 years ago

A 3kg object has an initial velocity (6i - 2j) m/s (a ) what is its kinetic energy at this time? (b) Find total work done on the

guapka [62]

Answer:

$K.E = \frac{1}{2} m {v}^{2} \\ {v}^{2}_i = {v}^{2} _x + {v}^{2} _y \\ = {(6)}^{2} + {( - 2)}^{2} = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i = \frac{1}{2} (3) (40) = 60J \\ \\ {v}^{2}_f = {v}^{2} _x + {v}^{2} _y \\ = {(8)}^{2} + {(4)}^{2} = 80m. {s}^{ - 1} \\ K.E_i = \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\ = 120J - 60J \\ = 60J$

3 0

3 years ago

A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How lo

gregori [183]

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

   H(t) = (H₀) + (v₀ T) + (1/2 a T²)

Height at any time 'T' after the drop =

(initial height) +
      (initial velocity) x (T) +
   (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

   Upward = the positive direction

   Initial height = +150 m
   Initial velocity = + 3 m/s

   Downward = the negative direction

   Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

   H(t) = (H₀) + (v₀ T) + (1/2 a T²)

   0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

   -4.9 T² + 3T + 150 = 0

Use the quadratic equation:

   T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]

   = (-1/9.8) [ -3 plus or minus 54.305 ]

   = (-1/9.8) [ 51.305 or -57.305 ]

      T = -5.235 seconds or 5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

H(t) = (H₀) + (v₀ T) + (1/2 a T²)

   H'(t) = v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set    v₀ + a T = 0

+3 - 9.8 T = 0

Add 9.8 to each side: 3   = 9.8 T

Divide each side by 9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh ! I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
   (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .

Distance climbed during that time = (average speed) x (time)

   = (1.5 m/s) x (0.306 sec)

   = 0.459 meter (hardly any at all)

   But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at 150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely 5.847 seconds after its release.

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work. The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

6 0

3 years ago