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3 years ago

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A sample of an unidentified compound contains 29.84% sodium, 67.49% chromium, and 72.67% oxygen. What is the compound's empirica

l formula? If the molar mass of the molecular formula is 523.96 g/mol, find the molecular formula.
Empirical Formula:
Molecular Formula:

1 answer:

vladimir1956 [14]3 years ago

5 0

Answer:

i would help i just dont know sorry

Explanation:

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When a 5.00-g metal piece, A, was immersed in 38.0 mL of water, the water level rose to 50.0 mL. Similarly, when a 5.00-g metal

Mnenie [13.5K]

Answer:

A is denser than B as it's volume for the same mass is smaller.

Explanation:

Hello.

In this case, we first need to take into account that the density of each metal A and B is computed by dividing the mass over the volume of each metal which is actually computed by substracting the volume of water from the volume of the water and the solid:

$V_A=50.0mL-38.0mL=12.0mL\\\\V_B=60.0mL-38.0mL=22.0mL$

Next, we compute the densities as shown below:

$\rho_A=\frac{m_A}{V_A}=\frac{5.00g}{12.0mL}=0.42g/mL\\ \\\rho_B=\frac{m_B}{V_B}=\frac{5.00g}{22.0mL}=0.23g/mL$

In such a way, A is denser is B as it's volume for the same mass is smaller.

Best regards.

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3 years ago

Examples of two substances being mixed and the

yKpoI14uk [10]

Answer:

ocean water liquid and steel solid

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3 years ago

Need help on question 3 plz<br> URGENT HELP PLZ

Law Incorporation [45]

Alloys are the homogeneous mixture of metals and non metals.... they are used in making cars and jewellery to make them more featurable,,, means that they would possess high regidity to the destruction .... both features of metal and nonmetal

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the student records the concentration of the stock solution of H2O2 to be 0.950 M. The student proceeds to prepare the reaction

lana [24]

Answer:

0.0400M of KI

Explanation:

Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.

When you add 10.0 mL of 0.10M KI and 15.0mL, total volume is:

25.0mL = <em>0.025L of solution</em>

<em />

And moles of KI are:

0.0100L × 0.10M = <em>0.00100 moles of KI</em>

<em />

Thus, molarity is:

0.00100 moles / 0.025L = <em>0.0400M of KI</em>

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3 years ago

Question 2 of 50

wolverine [178]

The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

Write the balanced chemical equation for the given reaction.
Convert the mass of calcium carbonate into moles.
Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

$CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}$

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

$mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}$

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

$mol \ CaO \ = 0.2498 \ mol$

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

$mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g$

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

$\boxed {mass \ CaO \ = 14 \ g}$

Learn More

Learn more about stoichiometry brainly.com/question/12979299
Learn more about mole conversion brainly.com/question/12972204
Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

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2 years ago