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Arlecino [84]
2 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]2 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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Here we have to choose the right option which tells the moles of CaCl₂ will react with 6.2 moles of AgNO₃ in the reaction

2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

6.2 moles of silver nitrate (AgNO₃) will react with B. 3.1 moles of calcium chloride (CaCl₂).

From the reaction: 2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂

Thus 2 moles of AgNO₃ reacts with 1 mole of CaCl₂

Henceforth, 6.2 moles of AgNO₃ reacts with \frac{6.2}{2} = 3.1 moles of CaCl₂.

1 mole of CaCl₂ reacts with 2 moles of AgNO₃. Thus-

A. 2.2 moles of CaCl₂ will react with 2.2×2 = 4.4 moles of AgNO₃.

C. 6.2 moles of CaCl₂ will reacts with 6.2×2 = 12.4 moles of AgNO₃.

D. 12.4 moles of CaCl₂ will reacts with 12.4 × 2 = 24.8 moles of AgNO₃

Thus the right answer is 6.2 moles of AgNO₃ will react with 3.1 moles of CaCl₂.

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Use the conversion factor 1amu=6.66054 x 10^-24 to answer the following questions.. . a) 1.674 x 10^-24 g of neutrons is how man
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A.) The conversion factor is 1 amu = 1.66054 ^{-24}g

To know how many amu in 1.674x10^{-24}g grams of neutrons:
1.674x10^{-24}g ( \frac{x (amu) }{ 1.66054x10^{-24} g)}

=1.00811 amu

b.) The mass in grams of one lithium ion which has an atomic weight of 6.94 amu.

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6.492x10^{-23}g ( \frac{x (amu) }{ 1.66054x10^{-24}g }) = 39.0957 amu


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2 years ago
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2 years ago
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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2(g) → 2SO3(g) At equilibrium, the partia
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Answer:

The partial pressure of SO₃ is 82.0 atm

Explanation:

The equilibrium constant Kp is equal to <em>the equilibrium pressure of the gaseous products raised to the power of their stoichiometric coefficients divided by the equilibrium pressure of the gaseous reactants raised to the power of their stoichiometric coefficients</em>.

For the reaction,

2 SO₂(g) + O₂(g) → 2 SO₃(g)

Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm

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