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3 years ago

6

give an example of a model used in science that is larger than the real object and an example of model that is smaller than the

real object

1 answer:

ad-work [718]3 years ago

4 0

A model of a atom and a model of the Earth

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A gamma ray photon has an energy of 4.75 x 10-14 joules. What is the frequency of this radiation?

Rama09 [41]

The frequency of the radiation is equal to $7.17 \times 10^{19}$ Hertz.

<u>Given the following data:</u>

Photon energy = $4.75 \times 10^{-14}$ Joules

To find the frequency of this radiation, we would use the Planck-Einstein equation.

Mathematically, the Planck-Einstein relation is given by the formula:

$E = hf$

<u>Where:</u>

h is Planck constant.

f is photon frequency.

Substituting the given parameters into the formula, we have;

$4.75 \times 10^{-14} = 6.626 \times 10^{-34} \times F\\\\F = \frac{4.75 \times 10^{-14}}{6.626 \times 10^{-34}}$

Frequency, F = $7.17 \times 10^{19}$ Hertz

Read more: brainly.com/question/16901506

4 0

3 years ago

During life body pH is ___ shortly after death the pH becomes ___ after time the pH becomes ___. (A) Acidic, neutral, basic (B)

Natalija [7]

(B) neutral, acidic, basic.

5 0

3 years ago

658 mL of 0.250 M HCl solution is mixed with 325 mL of 0.600 M HCl solution. What is the molarity of the resulting solution

Tju [1.3M]

The molarity of the resulting solution obtained by mixing 658 mL of 0.250 M HCl solution with 325 mL of 0.600 M HCl solution is 0.366 M

We'll begin by calculating the number of mole of HCl in each solution. This can be obtained as follow:

<h3>For solution 1:</h3>

Volume = 658 mL = 658 / 1000 = 0.658 L

Molarity = 0.250 M

<h3>Mole of HCl =?</h3>

Mole = Molarity x Volume

Mole of HCl = 0.250 × 0.658

<h3>Mole of HCl = 0.1645 mole</h3>

<h3>For solution 2:</h3>

Volume = 325 mL = 325 / 1000 = 0.325 L

Molarity = 0.6 M

<h3>Mole of HCl =?</h3>

Mole = Molarity x Volume

Mole of HCl = 0.6 × 0.325

<h3>Mole of HCl = 0.195 mole</h3>

Next, we shall determine the total mole of HCl in the final solution. This can be obtained as follow:

Mole of HCl in solution 1 = 0.1645 mole

Mole of HCl in solution 2 = 0.195 mole

Total mole = 0.1645 + 0.195

<h3>Total mole = 0.3595 mole</h3>

Next, we shall determine the total volume of the final solution.

Volume of solution 1 = 0.658 L

Volume of solution 2 = 0.325 L

Total Volume = 0.658 + 0.325

<h3>Total Volume = 0.983 L</h3>

Finally, we shall determine the molarity of the resulting solution.

Total mole = 0.3595 mole

Total Volume = 0.983 L

<h3>Molarity =?</h3>

Molarity = mole / Volume

Molarity = 0.3595 / 0.983

<h3>Molarity = 0.366 M</h3>

Therefore, the molarity of the resulting solution is 0.366 M

Learn more: brainly.com/question/25342554

3 0

3 years ago

If a 12-volt battery produces a current of 20 amps, what is the resistance?

mixas84 [53]

Answer:

0.6 Ω

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12 V

Current (I) = 20 A

Resistance (R) =?

From Ohm's law,

V = IR

Where:

V => is the voltage

I => is the current

R => R is the resistance

With the above formula, we can obtain the resistance as follow:

Voltage (V) = 12 V

Current (I) = 20 A

Resistance (R) =?

V = IR

12 = 20 × R

Divide both side by 20

R = 12 / 20

R = 0.6 Ω

Thus the resistance is 0.6 Ω

4 0

3 years ago

For 2,663 kg of a compound with the formula Al(SO), determine the following quantities (4 pts each); a) The number of moles of t

Nastasia [14]

Answer:

a) 35.485 moles of Al(SO)

b) 35.485 moles of S atoms

c) $2.136197(10^{25})$ Al atoms

d) 567.723 g of O

Explanation:

Let's define the following terms :

1 mol = $6.02.(10^{23})$ elemental units

For example :

1 mol of oxygen atoms = $6.02.(10^{23})$ oxygen atoms

Now, our compound has the following formula

Al(SO)

Where Al is aluminium

S is sulfur

And O is oxygen

All the subscripts are 1 so we can say the following :

1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O

In terms of moles :

1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O

The molar masses of Al, S and O are

$molarmass_{(Al)}=26.982\frac{g}{mol}$

$molarmass_{(S)}=32.065 \frac{g}{mol}$

$molarmass_{(O)}=15.999\frac{g}{mol}$

If we sum all the molar masses = $(26.982+32.065+15.999)\frac{g}{mol}=75.046\frac{g}{mol}$

Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.

1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.

Now we can calculate a),b),c) and d)

For a)

$2.663 kg=2663g$

75.046 g of Al(SO) = 1 mol of Al(SO)

2663 g of Al(SO) = x

$x=\frac{2663}{75.046}mol=35.485 mol$

2.663 kg of Al(SO) contains 35.485 moles of Al(SO)

b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms

We have 35.485 moles of Al(SO) molecules so

We have 35.485 moles of S atoms

And 35.485 moles of Al atoms

If 1 mol = $6.02(10^{23})$

35.485 moles of Al have $(35.485)(6.02)(10^{23})=2.136197(10^{25})$ Al atoms

d) 75.046 g of Al(SO) contains 15.999 g of O

2663 g of Al(SO) contains x g of O

$x=\frac{(2663).(15.999)}{75.046} g$

x = 567.723 g of O

6 0

3 years ago