Answer:
The original halide's formula is SrCl₂.
Explanation:
- The chemistry reaction's balanced equation is:
SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X indicates the halide.
- Based on the equation's stoichiometry, 1.0 mole of strontium halide yields 1.0 mole of SrSO₄.
- The moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
- The moles of SrX can thus be calculated as 4.11 x 10⁻³ moles based on stoichiometry from the balanced equation.
- n = mass / molar mass, thus n = 4.11 x 10⁻³ moles and mass = 0.652 g.
- The molar mass of SrX₂ is calculated using mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
- The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
- Calculating the atomic mass of halide X, we find = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.
- This identifies the atomic mass of Cl.
- Consequently, the original halide's formula is SrCl₂.
Explanation:
Initial moles of ethanoic acid = 0.020 mol
At equilibrium, half of the ethanoic acid molecules have reacted.
Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)
= 0.010 mol
Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol
The moles of product 
gas formed are determined as follows:
0.010 mol CH3COOH * (1 mol / 2 mol CH3COOH)
= 0.005 mol
Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol
Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol
Next, let’s determine the pressure:
0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:
P1/n1 = P2/n2
P2 = P1*(n2 / n1)
= 0.74 atm * (0.015 mol / 0.020 mol)
= 0.555 atm
First convert grams of C4H10 to moles using its molar mass of 58.1 g/mol:
3.50 g C4H10 × (1 mol C4H10 / 58.1 g C4H10) = 0.06024 mol C4H10
Next convert moles to molecules using Avogadro’s number:
0.06024 mol C4H10 × (6.022×10^23 molecules C4H10 / 1 mol C4H10) = 3.627×10^22 molecules C4H10
Each butane molecule contains 4 carbon atoms, so:
3.627×10^22 molecules C4H10 × (4 atoms C / 1 molecule C4H10) = 1.45×10^23 carbon atoms present.
Answer:Sugar-water is a mixture
Explanation:
If it consists of pure sugar, it's classified as neither; however, when mixed with water, it forms a homogeneous mixture.
The quantity of carbon atoms in the pencil mark amounts to 3 x 10^17. Given that the atomic weight of carbon is 12.01 amu, it follows that 12.01 g of carbon contains 6.022 x 10^23 atoms. Thus, we can set up the equation: 12.01 g carbon/ 6.022 x 10^23 atoms (3 x 10^17 atoms) (12.01 g carbon/ 6.022 x 10^23 atoms). By canceling out the atoms, we have (3 x 10^17) (12.01 g carbon/6.022 x 10^23) and then completing the division and multiplication yields 6 x 10^-6 g of carbon. Therefore, the mass of the pencil mark is 6 x 10^-6 g.
