A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

In a 100 g sample of the compound, there are 63.57 g of carbon, 6 g of hydrogen, 9.267 g of nitrogen, and 21.17 g of oxygen. First, convert these masses into moles (n) using the formula n = m/M, where M is the molar mass from the periodic table. For carbon: 63.57 g C -> 63.57 g C / 12.01 g/mol = 5.29 moles C. For hydrogen: 6 g H -> 6 g H / 1.008 g/mol = 5.95 moles H. For nitrogen: 9.267 g N -> 9.267 g N / 14.01 g/mol = 0.6615 moles N. For oxygen: 21.17 g O -> 21.17 g O / 16.00 g/mol = 1.32 moles O. Thus, the mole ratio looks like this: C 5.29 H 5.95 N 0.6615 O 1.32. Now, divide each value by the smallest number (1.32): C 4 H 4.5 N 0.5 O 1. To eliminate fractions, multiply all values by 2, yielding C8H9N1O2. Now, all numbers are integers! Hence, the empirical formula is C8H9NO2. Although the empirical formula isn't always the same as the molecular formula, in this instance, it corresponds to acetaminophen.

Option d is the correct choice, as both belong to the alkali metals category (group one).

To find the answer, start by calculating the total mass of the copper utilized:

Copper used = 100 pennies x 3.0g Cu per penny = 300.0 g Cu

Next, identify the path and molar ratios from Cu produced back to CuFeS2 needed using the established balanced reactions:

1 Cu2S from 2 CuS; 2Cu from 1 Cu2S; 2CuS from 2CuFeS2
Thus, 2Cu comes from 2CuFeS2, indicating a 1:1 molar ratio.

Then convert grams of Cu to moles and grams of CuFeS2:
= 300.0 g Cu * 1 mol Cu/63.546g Cu * 2 mol CuFeS2/2 moles Cu

= 4.72 moles CuFeS2

The required amount of chalcopyrite mined = 4.72 moles CuFeS2 * 183.54 g CuFeS2/1 mole CuFeS2 = 866.49 g CuFeS2

Step 1: Convert density from g/mL to g/L; 0.807 g/mL is equivalent to 807 g/L. Step 2: Calculate Moles of N₂; Density = Mass / Volume, or Mass = Density × Volume. Plugging in values, Mass = 807 g/L × 1 L gives us Mass = 807 g. Similarly, Moles = Mass / M.mass, which leads to Moles = 807 g / 28 g.mol⁻¹, giving us Moles = 28.82 moles. Step 3: Apply the Ideal Gas Law to determine Volume of gas occupied; P V = n R T, thus V = n R T / P. Remember to convert temperature to Kelvin (25 °C + 273 = 298 K). Hence, V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm, resulting in V = 704.76 L.

Answer:

0.400 moles of Oxygen

Explanation:

By employing the equation PV = nRT, the initial pressure of the flask can be calculated prior to the reaction, which leads to:

P = nRT/V

Where:

n signifies moles (4,000 moles: 2,000 moles of CO and 2,000 moles of H₂O)

R represents the gas constant (0.082 atm·L/mol·K)

T is the temperature (300.0 K)

V denotes volume (0.2000 L)

Substituting values results in P = 492.0 atm

To achieve a pressure reduction of 10.00%, the resulting pressure should be:

492.0 atm - 49.2 atm = 442.8 atm

Calculating with the new pressure under the same conditions gives the moles as:

n = PV/RT

n = 3,600 total moles

In the reaction:

2CO(g) + O₂(g) ⟶ 2CO₂(g)

The resulting moles are:

CO: 2,000 moles - 2X

O₂: 2,000 moles - X

CO₂: 2X

Where X accounts for the moles that react

Consequently, the total moles are:

4,000 moles - X = 3,600 moles

X = 0.400 moles

This indicates that the amount of oxygen needed for the reaction is 0.400 moles of Oxygen

I hope this is useful!