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Alex_Xolod [135]
3 years ago
10

You have 1 1/2 moles of 1 kg bottles of O2. What is the mass of O2 that you have? A. 9.033x10^23 kg B. 9.033x10^23 atoms C. 1.80

x10^27 kg D. 1.806x10^24 moles

Chemistry
2 answers:
ki77a [65]3 years ago
8 0

Answer:

B

Explanation:

Rashid [163]3 years ago
6 0

Answer:

b

Explanation:

trust

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Cytotoxic t cells can attack target cells with which chemical weapons?
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secrete cytotoxic substance which triggers apoptosis of target cell.

Explanation:

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4 0
3 years ago
How many moles of h+ are associated with the acid h2so3 during neutralization?
Katena32 [7]

H2SO3 or sulfurous acid is actually a strong acid. We know for a fact that strong acids completely dissociate into its component ions in a solution, that is:

 

<span>H2SO3 -->  2H+  +  SO3-</span>

 

<span>So from the equation above, there are 2 moles of H+</span>

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3 years ago
Select the options that are properties of electromagnetic waves:
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They propagate in materialistic media and non-materialistic media ( space ) .

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Explanation:

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4 0
3 years ago
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
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