What does internal combustion engines release?

How many moles are there in 2.30 grams of NH4SO2?

Anton [14]

10 gramas r in 2.30

6 0

3 years ago

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What is Dalton's law of partial pressure

Leto [7]

If 3 substances be x,y,z

$\\ \tt\longmapsto P_{net}=P_x+P_y+P_z$

Where

$\\ \tt\longmapsto P_x=\dfrac{n_xRT}{V}$

$\\ \tt\longmapsto P_y=\dfrac{n_yRT}{V}$

$\\ \tt\longmapsto P_z=\dfrac{n_zRT}{V}$

P is partial pressure
n is number of moles
V is volume
T is temperature

7 0

3 years ago

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Calculate the energy change when an electron moves from n=5 to n=7. Explain/show work please.

Korolek [52]

Answer: E = 1.55 ⋅ 10 − 19 J

Explanation:

The energy transition will be equal to 1.55 ⋅ 10 − 1 J .

So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

1 λ = R ⋅ ( 1 n 2 final − 1 n 2 initial ) , where λ - the wavelength of the emitted photon; R

- Rydberg's constant - 1.0974 ⋅ 10 7 m − 1 ; n final - the final energy level - in your case equal to 3; n initial - the initial energy level - in your case equal to 5. So, you've got all you need to solve for λ , so 1 λ =

1.0974 ⋅10 7 m − 1 ⋅ (.... −152

)

1

λ

=

0.07804

⋅

10

7

m

−

1

⇒

λ

=

1.28

⋅

10

−

6

m

Since

E

=

h

c

λ

, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by

h

⋅

c

, where

h

- Planck's constant -

6.626

⋅

10

−

34

J

⋅

s

c

- the speed of light -

299,792,458 m/s

So, the transition energy for your particular transition (which is part of the Paschen Series) is

E

=

6.626

⋅

10

−

34

J

⋅

s

⋅

299,792,458

m/s

1.28

⋅

10

−

6

m

E

=

1.55

⋅

10

−

19

J

8 0

3 years ago

In a single displacement reaction between sodium phosphate and barium, how much of each product (in grams) will be formed from 1

weeeeeb [17]

Answer:

A. 3.36g of Na.

B. 14.62g of Ba3(PO4)2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Ba + 2Na3PO4 → 6Na + Ba3(PO4)2

Next, we shall determine the mass of Ba that reacted and the mass of Na and Ba3(PO4)2 produced from the equation.

This is illustrated below:

Molar Mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 6 x 23 = 138g

Molar mass of Ba3(PO4)2 = (3 x 137) + 2[31 + (4x16)] = 411 + 2[31 + 64] = 601g/mol

Mass of Ba3(PO4)2 from the balanced equation = 1 x 601 = 601g

Summary:

From the balanced equation above,

411g of Ba reacted to produce 138g of Na and 601g of Ba3(PO4)2.

A. Determination of the mass of Na produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 138g of Na.

Therefore, 10g of Ba will react to produce = (10 x 138)/411 = 3.36g of Na.

Therefore, 3.36g of Na is produced.

B. Determination of the mass of Ba3(PO4)2 produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 601g of Ba3(PO4)2.

Therefore, 10g of Ba will react to produce = (10 x 601)/411 = 14.62g of Ba3(PO4)2.

Therefore, 14.62g of Ba3(PO4)2 is produced.

7 0

3 years ago

Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas phase as follows: C2H5I(g) → C2H4(g) + HI(g) From the follow

lutik1710 [3]

Answer:

Zero order

Explanation:

Looking at the data we can note a linear dependence between concentration and time.

Time Conc.

0 2

15 1.82

30 1.64

48 1.42

75 1.10

In the first 15 min it was consumed 2-1.82=0.18. So the rate is $r=\frac{\Delta C}{\Delta t} = \frac{0.18}{15}=0.012$

From 15 to 30 min (it has passed 15 min) is consumed 1.82-1.64=0.18, so as in the previous calculation the rate is $r=0.012$ .

From 30 to 48 (it has passed 18 min)the rate is $r= \frac{0.22}{18}\approx 0.012$

From 48 to 75 (it has passed 27 min) the rate is $r= \frac{0.32}{27}\approx 0.012$

So these results suggest that despite of the ever minor concentration of the reactant the rate is ever the same. Hence the reaction rate could be expressed as $r= k^{0} = 0.012 mol L^{-1} min^{-1}$ that is, the reaction is the zero order respect to C2H5I since it is not depending on concentration of C2H5I.

7 0

3 years ago