All categories

4 years ago

Empirical formula of a compound if a 50.0 g sample of it contains 9.1g Na ,20.6 g Cr ,and 22.2 g O

Chemistry

1 answer:

denpristay [2]4 years ago

5 0

Answer:

The empirical formula of a compound is Na2Cr2O7

Explanation:

Empirical formula is defined as the lowest ratio of elements present in any substance. In order to determine the empirical formula for a substance containing these elements we have to follow some steps.

1.In the first step, we have to divide the given mass of an element with its molecular mass to convert the mass to moles.

$No. of moles of Na=Given Mass of Na/Molecular Mass of Na=9.1/23=0.396$

Similarly, the moles of Cr and O in that compound were calculated and found as 0.396 and 1.3875, respectively.

2. Then the next step is to determine the lowest moles and divide all the moles with that lowest mole. As here the lowest mole is 0.396. So on dividing the moles of Na, Cr and O with 0.396, we will obtain the ratio as 1:1:3.5.

3. As the ratio consists of a fractional number, we have to multiply it with the smallest possible integer to convert it to whole number. In this case, if we multiply it with 2 , we will get the ratio as 2:2:7.

Thus, the empirical formula for the compound can be written as Na2Cr2O7.

You might be interested in

aluminum has a specific heat of 0.897 j/g°C. how much heat is needed to raise the temperature of a 79 gram piece of aluminum 28°

frutty [35]

Answer:

Approximately 2000 J.

General Formulas and Concepts:

Thermodynamics

Specific Heat Formula: q = mcΔT

q is heat (in J)
m is mass (in g)
c is specific heat (in J/g °C)
ΔT is change in temperature (in °C or K)

Explanation:

Step 1: Define

Identify variables

[Given] c = 0.897 J/g °C

[Given] m = 79 g

[Given] ΔT = 28°C

[Solve] q

Step 2: Solve for q

Substitute in variables [Specific Heat Formula]: q = (79 g)(0.897 J/g °C)(28 °C)
Multiply [Cancel out units]: q = (70.863 J/°C)(28 °C)
Multiply [Cancel out units]: q = 1984.16 J

Step 3: Check

Follow sig fig rules and round. We are given 2 sig figs as our lowest.

1984.16 J ≈ 2000 J

6 0

3 years ago

Hydrogen can reduce copper oxide but not aluminium oxide explain

Llana [10]

Answer: as copper has lower electrode potential value than hydrogen, it could be reduced by hydrogen.

Explanation: hydrogen has zero reduction potential while Cu has +0.34V and Al has -1.66 V .

SO in electrochemical series who has most negative or less reduction potential value tends to be a good reducing agent than the other.

Hope it helps...

3 0

3 years ago

Read 2 more answers

When MgCl2 dissolve in water does the Mg and Cl split apart?

Artyom0805 [142]

Answer:

Yes they do,

Explanation:

The spitting is called dissociation

7 0

3 years ago

Competition occurs when

Katen [24]

Answer:

Competition occurs when (A) two or more organisms need the same resource.

Explanation:

I dont think it could be (B), (C) or (D) because those don't really make sense.

3 0

2 years ago

In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe

AnnZ [28]

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

$\text{NH}_4\text{Cl}$ is a salt soluble in water. $\text{NH}_4\text{Cl}$ dissociates into ions completely when dissolved.

$\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq)$ .

The first test tube used to contain $\text{NH}_4\text{OH}$ . $\text{NH}_4\text{OH}$ is a weak base that dissociates partially in water.

$\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+} \;(aq)+ {\text{OH}}^{-} \; (aq)$ .

There's also an equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$ .

$\text{OH}^{-}$ ions from $\text{NH}_4\text{OH}$ will shift the equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ to the right and reduce the amount of ${\text{H}_3\text{O}}^{+}$ in the solution.

The indicator equilibrium will shift to the right to produce more ${\text{H}_3\text{O}}^{+}$ ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding $\text{NH}_4\text{Cl}$ will add to the concentration of ${\text{NH}_4}^{+}$ ions in the solution. Some of the ${\text{NH}_4}^{+}$ ions will combine with $\text{OH}^{-}$ ions to produce $\text{NH}_4\text{OH}$ .

The equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions will shift to the left to produce more of both ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$

The indicator equilibrium will shift to the left as the concentration of ${\text{H}_3\text{O}}^{+}$ increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

7 0

3 years ago