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kykrilka [37]
4 years ago
11

When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t

han ordinary seawater) is evaporated, 196 grams of MgCl2 are recovered. What is the molarity of MgCl2 in the original sample?
Chemistry
1 answer:
neonofarm [45]4 years ago
4 0

Answer:

2.06 M

Explanation:

number of moles of MgCl₂ = mass given in gram / molar mass

molar mass of MgCl₂ = 95.211 g/mol

number of moles of MgCl₂ = 196 g / 95.211 g/mol = 2.06 mol

Molarity of MgCl₂ in the original sample = number of moles / volume in Liters = 2.06 mol / 1 L = 2.06 M

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Calculate the number of moles represented by 21.7 g fe.
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The correct option is (e) 0.389 moles.

The number of moles represented by 21.7 g fe 0.389 moles.

Solution:

(21.7g) x (1mol/55.85g) = 0.389 moles

<h3>What is moles?</h3>

The mole, represented by the symbol mol, is the fundamental SI unit of material quantity. The amount of a substance determines how many elementary entities of that substance are present in an object or sample. Since the SI fundamental units were revised in 2019, the mole is now exactly defined as 6.022140761023.

<h3>Why the mole is a base unit?</h3>

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4 0
2 years ago
What is the chemical formula for lead IV oxide
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7 0
3 years ago
Which of the following species has the greatest
Vikki [24]

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Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr
Len [333]

Answer:

a. 1875 nm

b. 4051 nm

c. 1282 nm

These all are infrared electromagnetic radiation.

Explanation:

Our strategy here is to utilize the Rydberg equation for hydrogen atom electronic transition.

1/λ = Rh x (1/n₁² - 1/n₂²)   where  λ is the wavelength

                                                   Rh is Rydberg constant

                                                   n₁ and n ₂ are the energy levels ( n₁ < n₂ )

Now lets star the calculations.

a.  n₁  = 3, n₂ = 4

1/λ = 1.097 x 10⁷ /m x (1/3² - 1/4²) = 5.333 x 10⁵/m

λ  = 1/(5.333 x 10⁵ /m) = 1.875 x 10⁻⁶ m

Converting λ to nanometers:

1.875 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1875 nm

b.  n₁  = 4, n₂ = 5

1/λ = 1.097 x 10⁷ /m x  (1/4² - 1/5²) = 2.468 x 10⁵/m

λ  = 1/(2.468 x 10⁵/m) = 4.051 x 10⁻⁶ m

4.051 x 10⁻⁶ m x  (1 x10⁹ nm/m)  = 4051 nm

c.  n₁  = 3, n₂ = 5

1/λ = 1.097 x 10⁷ /m x  (1/3² - 1/5²) = 7801 x 10⁵/m

λ  = 1/(7801 x 10⁵/m) = 1282 x 10⁻⁶ m

1282 x 10⁻⁶ m x  (1 x10⁹ nm/m)  = 1282 nm

All of these transitions fall in the infrared region of the spectrum.

8 0
3 years ago
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