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2 years ago

In an experiment, 170.9 g of C2H4 was reacted with an excess of O2, 164.1 g of CO2 is produced.

Chemistry

1 answer:

jonny [76]2 years ago

7 0

Answer:

$Y=30.6\%$

Explanation:

Hello,

In this case, given the reaction, the molar mass of ethene is 28 g/mol and the molar mass of carbon dioxide is 44 g/mol. With that information we compute the theoretical yield considering a 1:2 molar ratio respectively between them:

$m_{CO_20}^{theoretical}=170.9gC_2H_4*\frac{1molC_2H_4}{28gC_2H_4}*\frac{2molCO_2}{1molC_2H_4} *\frac{44gCO_2}{1molCO_2} =537.1gCO_2$

Thus, we compute the percent yield with the given grams of carbon dioxide:

$Y=\frac{m_{CO_2}^{real}}{m_{CO_2}^{theoretical}}*100 \% =\frac{164.1g}{537.1gCO_2} *100 \%\\\\Y=30.6\%$

Regards.

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The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution i

bija089 [108]

Answer:

110ml

Explanation:

Using the dilution equation, C1V1 = C2V2

Where C1 is the initial concentration of solution

C2 is final concentration of solution

V1 is intital volume of solution

V2 is final volume of solution.

From the question , C1=6M, C2=0.5M, V1=10ml, V2=?

$V2 =\frac{C1V1}{C2}$

$V2 =\frac{10*6}{0.5}$

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volume of water added = final volume -initial volume

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3 0

2 years ago

To determine the concentration of ethanol in cognac a 5.00 mL sample of the cognac is diluted to 0.500 L. Analysis of the dilute

julia-pushkina [17]

Answer : The molar concentration of ethanol in the undiluted cognac is 8.44 M

Explanation :

Using neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molar concentration of undiluted cognac = ?

$M_2$ = molar concentration of diluted cognac = 0.0844 M

$V_1$ = volume of undiluted cognac = 5.00 mL = 0.005 L

$V_2$ = volume of diluted cognac = 0.500 L

Now put all the given values in the above law, we get molar concentration of ethanol in the undiluted cognac.

$M_1\times 0.005L=0.0844M\times 0.500L$

$M_1=8.44M$

Therefore, the molar concentration of ethanol in the undiluted cognac is 8.44 M

4 0

2 years ago

93.2 mL of a 2.03 M potassium fluoride (KF) solution

Marrrta [24]

Answer:

1.98 M

Explanation:

Given data

Initial volume (V₁): 93.2 mL

Initial concentration (C₁): 2.03 M

Volume of water added: 3.92 L

Step 1: Convert V₁ to liters

We will use the relationship 1 L = 1000 mL.

$93.2mL \times \frac{1L}{1000mL} = 0.0932 L$

Step 2: Calculate the final volume (V₂)

The final volume is the sum of the initial volume and the volume of water.

$V_2 = 0.0932L + 3.92 L = 4.01L$

Step 3: Calculate the final concentration (C₂)

We will use the dilution rule.

$C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M$

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2 years ago

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When hydrogen chloride ionizes in water, which reactant is the proton donor? chlorine which reactant is the proton acceptor? hyd

AlladinOne [14]

Let's write the reaction first.

HCl + H₂O ---> H₃O⁺ + Cl⁻

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1. The proton donor is HCl because it donates H+ to water which yields a hydronium ion, H₃O⁺.
2. The proton acceptor is water.

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2 years ago

Liquid nitrogen has a density of 0.807 g/ml at –195.8 °c. if 1.00 l of n2(l) is allowed to warm to 25°c at a pressure of 1.00 at

Flauer [41]

Step 1: Change density from g/mL to g/L;

0.807 g/mL = 807 g/L

Step 2: Find Moles of N₂;
As,
Density = Mass / Volume
Or,
Mass = Density × Volume

Putting Values,

Mass = 807 g/L × 1 L

Mass = 807 g
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Putting values,

Moles = 807 g / 28 g.mol⁻¹

Moles = 28.82 moles

Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,

As,
P V = n R T

V = n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)

V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm

V = 704.76 L

8 0

2 years ago