What does h2 and o2 combine

Mariana [72]

Answer:

$\boxed {\boxed {\sf 1.90 \times 10^{23} \ atoms \ Si}}$

Explanation:

We are asked to find how many atoms are in 8.88 grams of silicon.

<h3>1. Grams to Moles </h3>

First, we convert grams to moles. We use the molar mass or the mass of 1 mole of a substance. These values are found on the Periodic Table as they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up silicon's molar mass.

Si: 28.085 g/mol

We will convert using dimensional analysis. Set up a conversion factor with the molar mass.

$\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

We are converting 8.88 grams of silicon to moles, so we multiply by this value.

$8.88 \ g \ Si *\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

Flip the fraction so the units of grams of silicon cancel.

$8.88 \ g \ Si *\frac{1 \ mol \ Si} { 28.085 \ g \ Si}$

$8.88 *\frac{1 \ mol \ Si} { 28.085 }$

$\frac {8.88} { 28.085 } \ mol \ Si$

$0.316183015845 \ mol \ Si$

<h3>2. Moles to Atoms </h3>

Next, we convert moles to atoms. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of silicon.

Set up another conversion factor.

$\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

Multiply by the number of moles we calculated.

$0.316183015845\ mol \ Si *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

The units of moles of silicon cancel.

$0.316183015845 * \frac {6.022 \times 10^{23} \ atoms \ Si}{1}$

$0.316183015845 * {{6.022 \times 10^{23} \ atoms \ Si}$

$1.90405412 \times 10^{23} \ atoms \ Si$

<h3>3. Significant Figures</h3>

The original measurement of 8.88 grams has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 0 in the hundredth place.

$1.90 \times 10^{23} \ atoms \ Si$

<u>8.88 grams of silicon contains 1.90 ×10²³ atoms of silicon.</u>

6 0

3 years ago

Which has the highest boiling point .33 m NH3 or .10 m Na2SO4

Aleksandr-060686 [28]

Answer:

0.33 mol/kg NH₃

Explanation:

Data:

b(NH₃) = 0.33 mol/kg

b(Na₂SO₄) = 0.10 mol/ kg

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from a solute.

(a) For NH₃,

The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.

1 mol NH₃ ⟶ 1 mol particles

i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water

(b) For Na₂SO₄,

Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)

1 mol Na₂SO₄ ⟶ 3 mol particles

i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water

The NH₃ has more moles of particles, so it has the higher boiling point.

5 0

3 years ago

How is energy related to the change of state represented by the model?

Cerrena [4.2K]

Answer:

Atoms gain energy as a solid changes to a liquid. If atoms energy during a change of state, they are pulled together by attractive forces and become more organized.

4 0

3 years ago

Which of the following options would be the best for dissolving PbBr2(s)?

Nezavi [6.7K]

Answer:

2) Add a solution of NaBr

Explanation:

Lead (II) bromide is an inorganic powdery substance that has a solubility in water of 0.973 g/100 mL at 20°C. It is insoluble in alcohol but is soluble in alkali, ammonia, NaBr, and KBr

PbBr₂ is slightly soluble in ammonia, and it reacts with NaOH to produce Pb(OH)₂ and NaBr

Therefore, the best solution for dissolving PbBr₂(s) is NaBr

4 0

3 years ago

Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to

Delvig [45]

Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

7 0

3 years ago