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Vinil7 [7]
3 years ago
8

A sample of a compound containing C, O, and silver (Ag) weighed 1.372 g. On analysis, it was found to contain 0.288 g O and 0.97

4 g Ag. The molar mass of the compound was determined to be 308.8 g/mol. What is the molecular formula of the compound?
Chemistry
2 answers:
Triss [41]3 years ago
5 0

Answer:

The molecular formula is Ag2C2O4

Explanation:

Step 1: Data given

A compound contains C, O and Ag

The mass of the compound is 1.372 grams

Mass of O = 0.288 grams

Mass of Ag = 0.974 grams

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass Ag =107.87 g/mol

The molar mass of the compound = 308.8 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles O = 0.288 grams / 16.0 grams

Moles O = 0.018 moles

Moles Ag = 0.974 grams / 107.87 g/mol

Moles Ag = 0.00903 moles

Moles C = 0.11 grams / 12.01 g/mol

Moles C = 0.00916 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

O: 0.018/ 0.00903 = 2

C: 0.0.00916 / 0.00903 = 1

Ag: 0.00903/0.00903 =

The empirical formula is AgCO2

The molar mass of this empirical formula is 151.88

We have to multiply the empirical formula by n

n = 308.8/151.88 = 2

2* (AgCO2) = Ag2C2O4

The molecular formula is Ag2C2O4

Sauron [17]3 years ago
3 0

Answer:

The molecular formula for the compound is Ag₂C₂O₄, silver oxalate.

Explanation:

This is an easy excersise, we relate the moles of the elements in the compound with the molar mass of the total compound.

The sample weighs 1.372 g, with 0.288 g of O, 0.974 g of Ag and the rest, C

Therefore, 1.372g - 0.288 g - 0.974g = 0.112 g are C

We convert the mass to moles

0.288 g / 16g/mol = 0.018 moles of O

0.974 g / 107.87 g/mol = 0.00903 moles of Ag

0.112 g / 12 g/mol = 0.00933 moles of C

Now, the use of the molar mass of the compound. Let's prepare some rules of three:

1.372 g of compound contain 0.018 moles of O, 0.00903 moles of Ag and 0.00933 moles of C

Then, 308.8 g of compound (a mol) must contain:

(308.8 . 0.018) / 1.372 = 4 moles of O

(308.8 . 0.00903) / 1.372 = 2 moles of Ag

(308.8 . 0.00933) / 1.372 =  2 moles of C

The molecular formula for the compound is Ag₂C₂O₄, silver oxalate.

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Alex17521 [72]

Answer:

The answer to your question is 25 grams

Explanation:

Data

half-life = 5730 years

sample = 200 g

after 3 half-lives

Process

Calculate the amount of sample after one, two and three half-lives.

After each half-life, that of sample is half the previous amount.

                         Number of half-lives           Amount of sample

                                       0                                        200 g

                                        1                                         100 g

                                        2                                          50 g

                                       3                                           25 g

4 0
3 years ago
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When you receive a loan, the money the lender gives you is called the ____________. A. Interest B. Line of credit C. Principal D
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8 0
4 years ago
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If the energy of photon emitted from the hydrogen atom is 4.09 x 10-19 J, what is
Aliun [14]

Answer:

486 nm

Explanation:

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

Energy (E) = 4.09×10¯¹⁹ J

Wavelength (λ) =?

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

Energy (E) = 4.09×10¯¹⁹ J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

4.09×10¯¹⁹ = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 4.09×10¯¹⁹ / 6.63×10¯³⁴

f = 6.17×10¹⁴ Hz

Next, we shall determine the wavelength of the photon. This can be obtained as follow:

Frequency (f) = 6.17×10¹⁴ Hz

Velocity of photon (v) = 3×10⁸ m/s

Wavelength (λ) =?

v = λf

3×10⁸ = λ × 6.17×10¹⁴

Divide both side by 6.17×10¹⁴

λ = 3×10⁸ / 6.17×10¹⁴

λ = 4.86×10¯⁷ m

Finally, we shall convert 4.86×10¯⁷ m to nm. This can be obtained as follow:

1 m = 1×10⁹ nm

Therefore,

4.86×10¯⁷ m = 4.86×10¯⁷ m × 1×10⁹ nm / 1 m

4.86×10¯⁷ m = 486 nm

Therefore, the wavelength of the photon is 486 nm

7 0
3 years ago
Please help.
mojhsa [17]

Explanation: An element is represented in the form of _{A}^{Z}\textrm{X} where,

X = Symbol of the element

A = Atomic Mass of the element X

Z = Atomic Number of element X

Hence, For the element Justwondoricium,

Symbol = Jw

Atomic number = 120

Atomic Mass = 224

Now, Atomic number = Number of electrons = Number of Protons

Number of electrons = 120

Number of Protons = 120

and Atomic Mass = Number of neutrons + Number of protons

Number of neutrons can be calculated as we know atomic mass and number of neutrons, Putting the numbers we get

Number of neutrons = 224 - 120 = 104

The nearest noble gas to the element having atomic number 120 is Oganesson (Og), which has an atomic number of 118, so the next two electrons will be filled in the 8s orbital.

Electronic Configuration of Jw is [Og]8s^2

This electronic configuration lets us know about the location of the element in periodic table.

As the electron is entering the 8th shell, it belongs to the 8th period and as the last electron enters the s-orbital, it belongs to the S-block of the periodic table.

When the s-electrons are 1, it belongs to Group 1.

When the s-electrons are 2, it belongs to Group 2.

As this element has 2 s-electrons, it belongs to the Group 2.

8 0
4 years ago
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