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Lapatulllka [165]
3 years ago
9

A compound contains 74.2 g Na and 25.8 g O. Determine the Empirical Formula for this compound

Chemistry
2 answers:
Ksenya-84 [330]3 years ago
4 0

Answer:

The empirical formula for the compound is Na2O

Explanation:

Data obtained from the question include:

Sodium (Na) = 74.2g

Oxygen (O) = 25.8g

We can obtain the empirical formula for the compound as follow:

First, divide the above by their individual molar mass as shown below:

Na = 74.2/23 = 3.226

O = 25.8/16 = 1.613

Next, divide the above by the smallest number

Na = 3.226/1.613 = 2

O = 1.613/1.613 = 1

Therefore, the empirical formula is:

Na2O

vazorg [7]3 years ago
3 0

Answer:

The empirical formula is Na2O

Explanation:

Step 1: Data given

Mass of Na = 74.2 grams

Molar mass of Na = 22.99 g/mol

Mass of O = 25.8 grams

Molar mass of O = 16.0 g/mol

Step 2: Calculate moles Na

Moles Na = mass Na / molar mass Na

Moles Na = 74.2 grams / 22.99 g/mol

Moles Na = 3.23 moles

Step 3: Calculate moles O

Moles O = 25.8 grams / 16.0 g/mol

Moles O = 1.61 moles

Step 4: Calculate the mol ratio

We divide by the smallest amount of moles

Na: 3.23 moles / 1.61 moles = 2

O: 1.61 moles / 1.61 moles = 1

This means for every O atom we have 2 Na atoms

The empirical formula is Na2O

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Answer:

Ca has the greater Ionization Energy because the Trend is I/e decreases as you move down a group. Therefore, Ca has the greater I/e

Explanation:

3 0
3 years ago
How many moles of Al will be consumed when 0.400 mol of Al2O3 are produced in
Vladimir [108]

Answer:

C) 0.800 mol

Explanation:

  • 4Al + 3O₂ → 2Al₂O₃

In order to <u>convert from moles of Al₂O₃ into moles of Al</u>, we'll need to use<em> the stoichiometric coefficients</em>, using a conversion factor that has Al₂O₃ moles in the denominator and Al moles in the numerator:

  • 0.400 mol Al₂O₃ * \frac{4molAl}{2molAl_2O_3} = 0.800 mol Al

So the correct answer is option C).

7 0
3 years ago
Compounds formed from only non-metals consist of particles called
sleet_krkn [62]

Answer:

Compounds formed from non-metals consist of molecules.

Explanation:

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I hope I helped, please correct me if I'm wrong!

5 0
3 years ago
4.45 kcal of heat was added to increase the temperature of a sample of water from 23.0 °C to 57.8 °C. Calculate
Alona [7]

Answer:

m = 4450 g

Explanation:

Given data:

Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)

Initial temperature = 23.0°C

Final temperature = 57.8°C

Specific heat capacity of water = 1 cal/g.°C

Mass of water in gram = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 57.8°C - 23.0°C

ΔT = 34.8°C

4450 cal = m × 1 cal/g.°C × 34.8°C

m = 4450 cal / 1 cal/g

m = 4450 g

4 0
3 years ago
4 NH3 + 7 O2 → 4 NO2 + 6 H2O What is the mole ratio between oxygen and nitrogen dioxide? 7 moles to 6 moles 4 moles to 6 moles 7
Mashutka [201]

Answer:

7:4

Explanation:

O2 : NO2

7 : 4

hope this helps :)

8 0
3 years ago
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