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2 years ago

Is garlic powder a pure substance

Chemistry

1 answer:

CaHeK987 [17]2 years ago

8 0

Answer: most store-bought isnt

Explanation:

there typically a lot of addivitves

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A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2

sukhopar [10]

Answer:

a) $\triangle G^{0} = 7.31 kJ/mol$

b) $K_{-1} = 0.0594 m^{-1} s^{-1}$

Explanation:

Equation of reaction:

$2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)$

Initial pressure 3 1 0

Pressure change 2P 1P 2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

$K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }$

$K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}$

Standard free energy:

$\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol$

b) value of k−1 at 27 °C, i.e. 300K

$K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}$

$K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}$

$K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2} }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}$

6 0

3 years ago

Determine the molarity of 1.2 mol KCl in 1.1 L of a solution?

victus00 [196]

Answer:

1.1 M

General Formulas and Concepts:

Molarity = moles of solute / liters of solution

Explanation:

Step 1: Define variables

1.2 mol KCL

1.1 L of solution

M = unknown

Step 2: Solve for Molarity

Substitute: M = 1.2 mol/1.1 L
Evaluate: M = 1.09091

Step 3: Check

We are given 2 sig figs. Follow sig fig rules.

1.09091 M ≈ 1.1 M

7 0

2 years ago

What mass of oxygen is needed for the complete combustion of 4.60×10−3g of methane?

sp2606 [1]

First step in answering the question is to establish a balanced chemical reaction equation. More specifically, a combustion chemical equation.

CH4 + 2O2 ---> CO2 + 2H20

Then using dimension analysis:

$4.60*10^{-3} g CH4 ( \frac{moleCH4}{16 g CH4}) * ( \frac{2mole O2}{mole CH4}) * ( \frac{32 g O2}{mole O2} ) = 0.0184 g O_{2}$

7 0

3 years ago

What tool would you use to test for hardness?

s344n2d4d5 [400]

Answer: mohs picks

Explanation: hope its what you wanted

8 0

2 years ago

Which of the following is NOT a good technique for managing used oil? A) Have specific, labeled catch pans available for technic

slava [35]

The answer is: B) Spills in your shop and any releases on pavement or outside should be poured down a drain.

Pouring used oil down a drain can cause damege to environment.

Oil is nonpolar substance and does not dissolve in the water (polar molecule), used oil remain existent in the drain and sewer or in the environment.

Small spills create a cumulative effect with significant impact on the environment.

Untended spills inside a shop can spread to an inside drain connected to the storm sewer system and cause floods.

8 0

2 years ago