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3 years ago

calculate the grams of oxygen gas present in a 2.50 L sample kept at 1.66 atm pressure and a temperature of 10.0 C

Chemistry

2 answers:

Ahat [919]3 years ago

8 0

Answer:

5.72 grams of O₂ are present in the sample

Explanation:

Let's calculate the moles of O₂ by the Ideal Gases Law.

P . V = n . R . T

1.66 atm . 2.50L = n . 0.082 . 283K

(1.66 atm . 2.50L) / 0.082 . 283K = n

0.179 mol = n

Molar mass O₂ = 32 g/m

Mol . molar mass = mass → 0.179 m . 32g/m = 5.72 g

Aloiza [94]3 years ago

7 0

Answer:

The mass of oxygen gas is 5.72 grams

Explanation:

Step 1: Data given

Volume of oxygen gas (O2) = 2.50L

Pressure of O2 gas = 1.66 atm

Temperature = 10.0 °C = 283 Kelvin

Step 2: Calculate moles of O2 gas

p*V = n*R*T

n = (p*V)/(R*T)

⇒ with n = the moles of O2 gas = TO BE DETERMINED

⇒ with p = the pressure of the O2 gas =1.66 atm

⇒ with V = the volume of the gas = 2.50 L

⇒ with R = the gas constat = 0.08206 L*atm/K*mol

⇒ with T= the temperature = 10.0°C = 283 K

n = (1.66 * 2.50)/(0.08206*283)

n = 0.1787 moles O2

Step 3: Calculate mass of O2 grams

Mass O2 = moles O2 = Molar mass O2

Mass O2 = 0.1787 moles * 32.0 g/mol

Mass O2 = 5.72 grams

The mass of oxygen gas is 5.72 grams

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2 years ago

A student needs to prepare 100. mL of 0.612 M Cu(NO3)2 solution. What mass, in grams, of copper(II) nitrate should the student u

Temka [501]

Answer: 11.5 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution

$Molarity=\frac{n\times 1000}{V_s}$

where,

Morality = 0.612 M

n= moles of solute

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Now put all the given values in the formula of molarity, we get

$0.612=\frac{n\times 1000}{100ml}$

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3 0

3 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

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3 years ago

Does ionic or covalent form solid crystals?

Alekssandra [29.7K]

Answer:

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3 years ago

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r

weqwewe [10]

Answer:

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by $\bigtriangledown H^\circ_{rxn}$

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ =?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1) $\bigtriangledown H^\circ_{rxn}$ = +157KJ

P₄(g)+6Cl₂(g)→ 4PCl₃(g).............(2) $\bigtriangledown H^\circ_{rxn}$ = -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3) $\bigtriangledown H^\circ_{rxn}$ = -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4) $\bigtriangledown H^\circ_{rxn}$ =4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ =( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ = - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

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3 years ago