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BlackZzzverrR [31]
3 years ago
14

Sanjay is learning about the elements that make up Earth’s crust. Ms. Richards, his Earth science teacher, gives him an element

and asks him to determine which element from Earth’s crust it is. She also provides Sanjay with this table. Sanjay weighs the element and finds that it has a mass of 17.98 g. By placing the element in water, he determines the volume of the substance to be 2.2835 mL.
Based on the table from Ms. Richards, what element from Earth’s crust did Ms. Richards gave to Sanjay?
A) iron
B) calcium
C) aluminum
D) silicon
Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer:

Iron

Hope this helps! Stay safe!

(Plz vote me as brainliest)

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Under what circumstances may a health insurer charge a higher premium to a woman with a genetic disposition to breast cancer? a)
wel

Answer:

D

Explanation:

No company is allowed to discriminate.

8 0
2 years ago
What is the energy of a wave with wavelength of 4.2 x 10-7 m. (Hint: Calculate for frequency first.)
erik [133]

Answer:

Option B. 4.74×10¯¹⁹ J.

Explanation:

The following data were obtained from the question:

Wavelength (λ) = 4.2×10¯⁷ m

Energy (E) =.?

Next, we shall determine the frequency of the wave. This can be obtained as follow:

Wavelength (λ) = 4.2×10¯⁷ m

Velocity (v) = constant = 3×10⁸ m/s

Frequency (f) =.?

v = λf

3×10⁸ = 4.2×10¯⁷ × f

Divide both side by 4.2×10¯⁷

f = 3×10⁸ / 4.2×10¯⁷

f = 7.143×10¹⁴ s¯¹

Therefore, the frequency of the wave is 7.143×10¹⁴ s¯¹.

Finally, we shall determine the energy of the wave using the following formula

E = hf

Where

E is the energy.

h is the Planck's constant

f is the frequency

Thus, the enery of the wave can be obtained as follow:

Frequency (f) = 7.143×10¹⁴ s¯¹.

Planck's constant = 6.63×10¯³⁴ Js

Energy (E) =..?

E = hf

E = 6.63×10¯³⁴ × 7.14×10¹⁴

E = 4.74×10¯¹⁹ J

Therefore, the energy of the wave is 4.74×10¯¹⁹ J.

5 0
3 years ago
Why is smoke filling up a room diffusion or not diffusion
miv72 [106K]
It is diffistion because it moves from an area of high concentration to low
8 0
3 years ago
The equilibrium constant for the reaction
FinnZ [79.3K]

Answer: The concentrations of Cl_2 at equilibrium is 0.023 M

Explanation:

Moles of  Cl_2 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol

Volume of solution = 1 L

Initial concentration of Cl_2 = \frac{0.14mol}{1L}=0.14M

The given balanced equilibrium reaction is,

                            COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial conc.           0.14 M           0 M       0M    

At eqm. conc.     (0.14-x) M        (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}

Now put all the given values in this expression, we get :

4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of Cl_2 at equilibrium is 0.023 M

7 0
3 years ago
1.24 moles of magnesium arsenate are dissolved in 1.74 kg of solution. Calculate the molality of the solution.
aleksklad [387]

Answer:

Molality of the solution = 0.7294 M

Explanation:

Given:

Number of magnesium arsenate = 1.24 moles

Mass of solution = 1.74 kg

Find:

Molality of the solution

Computation:

Molality of the solution = Mole of solute / Mass of solution = 1.74 kg

Molality of the solution = 1.24 / 1.7

Molality of the solution = 0.7294 M

6 0
3 years ago
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