Can someone tell me if I put this in the correct pairs???? Help please!!??

A hypothetical element consists of four isotopes. Use the information below to calculate the average atomic volume mass of the e

Keith_Richards [23]

Answer:

104.969 amu.

Explanation:

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

Isotope A:

Mass of A = 107.977 amu

Abundance (A%) = 0.1620%

Isotope B:

Mass of B = 106.976 amu

Abundance (B%) = 1.568%

Isotope C:

Mass of C = 105.974 amu

Abundance (C%) = 47.14%

Isotope D:

Mass of D = 103.973 amu

Abundance (D%) = 51.13%

Average atomic mass =?

The average atomic mass of the element can be obtained as follow:

Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]

Average atomic mass = [(107.977 × 0.1620)/100] + [(106.976 × 1.568)/100] + [(105.974 × 47.14)/100] + [(103.973 × 51.13)/100]

= 0.175 + 1.677 + 49.956 + 53.161

= 104.969 amu

Therefore, the average atomic mass of the element is 104.969 amu.

8 0

4 years ago

The mass number of 3 isotopes of magnesium are 24,25,26.what is the mass number of the most abundant isotope of magnesium.

Anarel [89]

Answer:

24.4 amu or g/mole

Explanation:

24 x 0.790 = 19.0 amu

25 x 0.100 = 2.50 amu

26 x 0.110 = 2.86 amu

(Because of the 19.0, the sig figs go only to the 1/10 decimal place)

19.0 + 2.5 + 2.9 = 24.4 amu or g/mole

7 0

3 years ago

How much<br>much hydrogen gas evolved<br>when 1.5 current is passed through water for 1.5 hours?

evablogger [386]

0.042 moles of Hydrogen evolved

<h3>Further explanation</h3>

Given

I = 1.5 A

t = 1.5 hr = 5400 s

Required

Number of Hydrogen evolved

Solution

Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.

Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)

Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻

Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)

So at the cathode H₂ gas is produced

Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C

$\tt mol~e^-=\dfrac{Q}{96500}$

Q = i.t

Q = 1.5 x 5400

Q = 8100 C

mol e⁻ = 8100 : 96500 = 0.084

From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042

4 0

3 years ago

Which is true about relative dating?

AleksAgata [21]

it's the last one because it's depended on the rock layers

8 0

3 years ago

Read 2 more answers

A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa

Liula [17]

Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

4 0

4 years ago