Volcano formation is directly related to plate tectonics. <br><br><br> True<br> False

How many molecules of sodium oxide will be created if 187 grams of oxygen reacts with excess sodium? 4Na + O2 -> 2Na2O A. 11.

Jet001 [13]

Answer:

<u>7.04 × 10²⁴ molecules of Na₂O</u>

Explanation:

<u>1) Balanced chemical equation (given):</u>

4 Na(s) + O₂ (g) → 2Na₂O (s)

<u>2) Mole ratio:</u>

1 mol O₂ : 2 mol Na₂O

<u>3) Number of moles of O₂ in 187 grams of oxygen:</u>

Molar mass O₂ = 32.00 g/mol

Number of moles = mass in grams / molar mass = 187 g / 32.00 g/mol = 5.844 mol

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<u>4) Number of moles of Na₂O:</u>

Proportion: 1 mol O₂ / 2 mol Na₂O = 5.844 mol O₂ / x

x = 5.844 mol O₂ (2 mol Na₂O / 1 mol O₂) = 11.688 mol Na₂O

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<u>5) Number of molecules</u>

Number of molecules = number of moles × Avogadro's number

Number of molecules = 11.688 mol × 6.022 × 10²³ molecules/mol =

= 7.04 × 10 ²⁴ molecules ← answer

The answer is rounded to three significant figures.

4 0

3 years ago

1. A sample of gold (Au) has a mass of 35.12 g.

Alecsey [184]

1a) Answer is: the number of moles of gold (Au) in the sample is 0.178 mol.

m(Au) = 35.12 g; mass of gold.

M(Au) = 196.97 g/mol; molar mass of gold.

n(Au) = m(Au) ÷ M(Au).

n(Au) = 35.12 g ÷ 196.97 g/mol.

n(Au) = 0.178 mol; amount of gold.

1b) Answer is: the number of atoms of gold (Au) is 1.073·10²³.

N(Au) = n(Au) · Na.

N(Au) = 0.178 mol · 6.022·10²³ 1/mol.

N(Au) = 1.073·10²³.

2a) Answer is: 0.0035 moles of sucrose.

m(C₁₂H₂₂O₁₁) = 1.202 g; mass of sucrose.

M(C₁₂H₂₂O₁₁) = 12 · Ar(C) + 22 · Ar(H) + 11 · Ar(O) · g/mol.

M(C₁₂H₂₂O₁₁) = 12 · 12.01 + 22 · 1.01 + 11 · 16 · g/mol.

M(C₁₂H₂₂O₁₁) = 342.3 g/mol; molar mass of sucrose.

n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).

n(C₁₂H₂₂O₁₁) = 1.202 g ÷ 342.3 g/mol.

n(C₁₂H₂₂O₁₁) = 0.0035 mol; amount of sucrose.

2b) n(C) = 12·n(C₁₂H₂₂O₁₁).

n(C) = 12 · 0.0035 mol.

n(C) = 0.042 mol; amount of carbon in sucrose.

n(H) = 22·n(C₁₂H₂₂O₁₁).

n(H) = 22 · 0.0035 mol.

n(H) = 0.077 mol; amount of hydrogen in sucrose.

n(O) = 11·n(C₁₂H₂₂O₁₁).

n(O) = 11 · 0.0035 mol.

n(O) = 0.0385 mol; amount of oxygen atoms in sucrose.

2c) N(C) = n(C) · Na.

N(C) = 0.042 mol · 6.022·10²³ 1/mol.

N(C) = 2.53·10²²; number of carbon atoms in sucrose.

N(H) = n(H) · Na.

N(H) = 0.077 mol · 6.022·10²³ 1/mol.

N(H) = 4.63·10²²; number of hydrogen atoms in sucrose.

N(O) = n(O) · Na.

N(O) = 0.0385 mol · 6.022·10²³ 1/mol.

N(O) = 2.31·10²²; number of oxygen atoms in sucrose.

Na is Avogadro constant.

4 0

3 years ago

Find the volume occupied by 4.00 g of helium gas at <br> 25.0o C and 1.18 atm of pressure.

Triss [41]

Answer: 41.46 L

Explanation:

La ecuación que describe relación entre presión, volumen, temperatura y la cantidad (en moles)

de un gas ideal es: PV = nRT

Donde: P = Presión absoluta , V= Volumen , n = Moles de gas , R = Constante universal de los gases ideales, T = Temperatura absoluta,

R = 0.082 L. atm/mol. °K

V = nRT/P

Calculanting n

n = mass/ molecular mass

6 0

3 years ago

Which one from the drop down box

mel-nik [20]

CH4+ 2O2---> CO2+ 2H2O I believe would be correct.

7 0

4 years ago

How many kilograms of a fertilizer are made of pure P2O5 would be required to supply 1.69 kilogram of phosphorus to the soil?

Alekssandra [29.7K]

Answer:

3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.

Explanation:

The molecular mass of P2O5 is

P2 = 2* 31 = 62

O5 = 5 *<u> 16 = 80</u>

Molecular Mass = 142

Set up a Proportion

142 grams P2O5 supplies 62 grams of phosphorus

x kg P2O5 supplies 1.69 kg of phosphorus

Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.

142/x = 62/1.69 Cross multiply

142 * 1.69 = 62x combine the left

239.98 = 62x Divide by 62

239.98/62 = x

3.89 kg of P2O5 must be used.

3 0

3 years ago

Volcano formation is directly related to plate tectonics. True False