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3 years ago

How many grams of water are made from the reaction of 16.0 grams of oxygen gas?

1 answer:

4 0

Assuming hydrogen gas is in excess, calculate the moles of oxygen

n(O2) = 16.0 g / 32.0 g/mol = 0.500 mol 

Calculate the moles of water using the mole ratio; the ratio of H2O to O2 is 2:1 or the moles of H2O is twice the moles of O2 

n(H2O) = 2 x n(O2) = (2)(0.500) = 1.00 mol 

Calculate the mass of H2O by multiplying the moles by the molar mass of water 

mass = (1.00 mol)(18.0 g/mol) = 18.0 g H2O 

Ans: A)

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11. Why is HCl a strong acid and HClO a weak acid?

ankoles [38]

Strong acid:dissolves and dissociates 1005 to produce protons (H+) 1. seven
strong acids: HCI, HBr, HI, HNO3, H2SO4, and HCIO3. ...
weak acid: dissolves but less than 100% dissociates to produce protons (H+) 1.

3 0

3 years ago

sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5. V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a) ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0

3 years ago

PLS HELP THE QUESTION IS ON THE PICTURE

IceJOKER [234]

Concepts used:

1 mole of an element or a compound has 6.022 * 10²³ formula units

So, we can say that: Number of formula units = number of moles * 6.022*10²³

number of moles of an element or a compound = given mass/molar mass

__________________________________________________________

003 - Number of CaH₂ formula units in 6.065 grams

Number of Moles:

We know that the molar mass of CaH₂ is 42 grams/mol

Number of Moles of CaH₂ = given mass/molar mass

Number of moles = 6.065 / 42

Number of moles = 0.143 moles

Number of Formula units:

Number of formula units = number of moles * 6.022*10²³

= 0.143 * 6.022 * 10²³

= 0.86 * 10²³ formula units

__________________________________________________________

004 - Mass of 6.34 * 10²⁴ formula units of NaBF₄

Number of Moles:

We mentioned this formula before:

Number of formula units = number of moles * 6.022*10²³

Solving it for number of moles, we get:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variable

Number of moles = 6.34 * 10²⁴ / 6.022*10²³

Number of moles= 10.5 moles

Mass of 10.5 moles of NaBF₄:

Molar mass of NaBF₄ = 38 grams/mol

Mass of 10.5 moles = 10.5 * molar mass

Mass of 10.5 moles = 10.5 * 38

Mass = 399 grams

__________________________________________________________

005 - Number of moles in 9.78 * 10²¹ formula units of CeI₃

Number of Moles:

We have the formula:

Number of moles = Number of Formula units / 6.022* 10²³

replacing the variables

Number of Moles = 9.78 * 10²¹ / 6.022*10²³

Number of Moles = 1.6 / 10²

Number of Moles = 1.6 * 10⁻² moles OR 0.016 moles

3 0

2 years ago

What happens when water is added to quicklime? Write two observations.

Neporo4naja [7]

Answer:

1.Most metal oxides are insoluble in water but some of these (e.g. Na2O.

Explanation:

2.: (i) A hissing sound is observed.

1.ii) The mixture starts boiling and lime water is obtained.

3 0

3 years ago

Compounds like CCl₂F₂ are known as chlorofluorocarbons, or CFCs. These compounds were once widely used as refrigerants but are n

natta225 [31]

Answer: Mass Of CFC that needs to evaporate for the freezing of water = 328.24 g

Explanation: Heat gained by the CFC = Heat lost by water

Heat lost by water = Heat required to take water's temperature to 0°c + Heat required to freeze water at 0°c

Heat required to take water's temperature from 33°c to 0°c = mCΔT

m = 201g, C = 4.18 J/(gK), ΔT = 33

mCΔT = 201 × 4.18 × 33 = 27725.94 J

Heat required to freeze water at 0°c = mL

m = 201g, L = 334 J/g

mL = 201 × 334 = 67134 J

Heat gained by CFC to vaporize = mH = 27725.94 + 67134 = 94859.94 J

H = 289 J/g, m = ?

m × 289 = 94859.9

m = 328.24 g

QED!!

7 0

3 years ago