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Arlecino [84]
3 years ago
12

A chromium oxide compound contains 104.0 grams of chromium and 48.0 grams of oxygen. What is the most likely empirical formula o

f this compound?
Chemistry
1 answer:
TEA [102]3 years ago
4 0

Answer: Empirical formula of this compound is Cr_2O_3

Explanation:

Mass of Cr= 104.0 g

Mass of O = 48.0 g

Step 1 : convert given masses into moles.

Moles of Cr =\frac{\text{ given mass of Cr}}{\text{ molar mass of Cr}}= \frac{104.0g}{52g/mole}=2moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{48.0g}{16g/mole}=3moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Cr = \frac{2}{2}=1

For O =\frac{3}{2}=1.5

Converting into simple whole number ratios by multiplying by 2

The ratio of Cr : O= 2: 3

Hence the empirical formula is Cr_2O_3

The most likely empirical formula of this compound is Cr_2O_3

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Explanation:

Because it have 2 cluster molecules

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Hatshy [7]

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4 0
10 months ago
What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0
3 years ago
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Pepsi [2]

Answer:

0 net force it is balanced

Explanation:

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7 0
3 years ago
What is the pH of a solution with (H+) =2.3x10 -6
castortr0y [4]
P
H
=
−
log
10
[
H
3
O
+
]
=
−
log
10
{
2.3
×
10
−
6
}
=
−
{
−
5.64
}
=
5.64
5 0
1 year ago
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