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2 years ago

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Acetylene gas (C2H2), is used in welding torches. When it reacts with oxygen, it produces carbon dioxide (CO2), and steam (H2O).

The reaction can be described by the equation 2C2H2+5O2 -> 4CO,+2H2O. How much mass of C2H2 is needed to produce 75.0 grams of CO2?

1 answer:

Anika [276]2 years ago

8 0

Answer:

22.1 g

Explanation:

Step 1: Write the balanced equation for the combustion of acetylene

2 C₂H₂ + 5 O₂ ⇒ 4 CO₂ + 2 H₂O

Step 2: Calculate the moles corresponding to 75.0 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

75.0 g × 1 mol/44.01 g = 1.70 mol

Step 3: Calculate the moles of C₂H₂ required to produce 1.70 moles of CO₂

The molar ratio of C₂H₂ to CO₂ is 2:4. The moles of C₂H₂ required are 2/4 × 1.70 mol = 0.850 mol.

Step 4: Calculate the mass corresponding to 0.850 moles of C₂H₂

The molar mass of C₂H₂ is 26.04 g/mol.

0.850 mol × 26.04 g/mol = 22.1 g

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Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

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2 years ago

At a temperature of __________ °c, 0.444 mol of co gas occupies 11.8 l at 889 torr.

Novay_Z [31]

<span>ideal gas law is: PV = nRT P = pressure (torr) = 889 torr V = volume (Liters) = 11.8 L n = moles of gas = 0.444 mol R = gas constant = 62.4 (L * torr / mol * k) solve for T (in kelvin) T = PV/nR T = (889*11.8)/(.444*62.4) T = 378.6 K convert to C (subtract 273) T = 105.6 deg C</span>

3 0

2 years ago

Read 2 more answers

Anna [14]

Answer:

just answer this and you will have yours

Explanation:Find the area of a circle with a diameter of \color{green}{16}16start color green, 16, end color green.

Either enter an exact answer in terms of \piπpi or use 3.143.143, point, 14 for \piπpi and enter your answer as a decimal.

8 0

1 year ago

Exactly 500 grams of ice are melted at a temperature of 32°f. (lice = 333 j/g.) calculate the change in entropy (in j/k). (give

denpristay [2]

Entropy Change is calculated by (Energy transferred) / (Temperature in kelvin)
deltaS = Q / T

Q = (mass)(latent heat of fusion)
Q = m(hfusion)
Q = (500g)(333J/g) = 166,500J

T(K) = 32 + 273.15 = 305.15K
deltaS = 166,500J / 305.15K
deltaS = 545.63 J/K

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2 years ago

A 0.20 mol sample of MgCl2(s) and a 0.10 mol sample of KCl(s) are dissolved in water and diluted to 500 mL. What is the concentr

igor_vitrenko [27]

Answer:

1 M

Explanation:

Magnesium chloride will furnish chloride ions as:

$MgCl_2\rightarrow Mg^{2+}+2Cl^-$

Given :

Moles of magnesium chloride = 0.20 mol

Thus, moles of chlorine furnished by magnesium chloride is twice the moles of magnesium chloride as shown below:

$Moles =2\times 0.20\ moles$

Moles of chloride ions by magnesium chloride = 0.40 moles

Potassium chloride will furnish chloride ions as:

$KCl\rightarrow K^{+}+Cl^-$

Given :

Moles of potassium chloride = 0.10 moles

Thus, moles of chlorine furnished by potassium chloride is same as the moles of potassium chloride as shown below:

Moles of chloride ions by potassium chloride = 0.10 moles

Total moles = 0.40 + 0.10 moles = 0.50 moles

Given, Volume = 500 mL = 0.5 L (1 mL = 10⁻³ L)

Concentration of chloride ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{Cl^-}=\frac{0.50}{0.5}$

<u> The final concentration of chloride anion = 1 M</u>

8 0

2 years ago