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Triss [41]
2 years ago
13

Uifuiiiddiixkxkdkddkdi

Chemistry
1 answer:
pantera1 [17]2 years ago
3 0

Answer:

haksonnehejskznnd

Explanation:

jdudubebs sbit

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What is the total number of hydrogen atoms contained in one molecule of (NH4)2C204?
Ber [7]

Answer:

=8 atoms

Explanation:

In (NH4)2C2O4 there are four moles of Hydrogen in the compound (NH4), but there two molecules of (NH4) in this compound.  That's what the 2 in (NH4)2 means, so multiply 4 x 2 = 8.

Hope this helps (:

8 0
3 years ago
Read 2 more answers
Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)
Fynjy0 [20]

Answer:

Rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

Explanation:

According to equation   2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

                     ⇒ -\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}  -----------------------------(1)

    Given that the rate of disappearance of oxygen = -\frac{d[O_{2} ]}{dt} = 3.64 x 10⁻³ M/s

             So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = ?

from equation (1) we can write

                                   \frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]

                                ⇒ \frac{d[SO_{3}] }{dt} = 2 x 3.64 x 10⁻³ M/s

                                ⇒ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

7 0
3 years ago
A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio
kondor19780726 [428]

Answer:

The final volume of the cylinder is 1.67 L

Explanation:

Step 1: Data given

Initial volume = 0.250 L

external pressure = 2.00 atm

Expansion does 288 J of work on the surroundings

Step 2: Definition of reversible work:

Wrev = -P(V2-V1) = -288 J

The gas did work, so V2>V1  (volume expands) and the work has a negative sign.(Wrev<0)

V2 = (-Wrev/P)  + V1

⇒ with Wrev = reverse work (in J)

⇒ with P = the external pressure (in atm)

⇒ with V1 = the initial volume

We can see that your pressure is in  atm  and energy in J

To convert from J to L * atm we should use a convenient conversion unit using the universal gas constants :

R = 8.314472 J/mol *K and R= 0.08206 L*atm/K*mol

V2 =- (-288 J * (0.08206 L*atm/K*mol  /8.314 J/mol *K))/2.00 atm  + 0.250L

V2 = 1.67 L

The final volume of the cylinder is 1.67 L

8 0
3 years ago
Question 3 help me asap please
e-lub [12.9K]

Its chemical formula H2O, indicates that each of its molecules contains one oxygen and two hydrogen atoms, connected by covalent bonds. The hydrogen atoms are attached to the oxygen atom at an angle of 104.45°. "Water" is the name of the liquid state of H2O at standard conditions for temperature and pressure.

5 0
2 years ago
Step I of the proposed mechanism involves the collision between NO2 and F2 molecules. This step is slow even though such collisi
pogonyaev

Answer:

bob

Explanation:

bob

7 0
3 years ago
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