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Margaret [11]
2 years ago
7

3) Rubidium reacts with rubidium nitrate to produce rubidium oxide and nitrogen gas

Chemistry
1 answer:
postnew [5]2 years ago
5 0

Answer:

3) Rubidium reacts with rubidium nitrate to produce rubidium oxide and nitrogen gas

a- Write the chemical reaction

10Rb + 2RbNO3 → 6Rb2O + N2

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in example 5.11 of the text the molar volume of n2 at STP is given as 22.42 L/mol how is this number calculatd how does the mola
Valentin [98]

Answer:

V = 22.42 L/mol

N₂ and H₂ Same molar Volume at STP

Explanation:

Data Given:

molar volume of N₂ at STP = 22.42 L/mol

Calculation of molar volume of N₂ at STP  = ?

Comparison of molar volume of H₂ and N₂ = ?

Solution:

Molar Volume of Gas:

The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol

Molar volume can be calculated by using ideal gas formula  

                               PV = nRT

Rearrange the equation for Volume

                            V = nRT / P . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

Standard values

P = 1 atm

T = 273 K

n = 1 mole

R = 0.08206 L.atm / mol. K

Now put the value in formula (1) to calculate volume for 1 mole of N₂

                   V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm

                   V = 22.42 L/mol

Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.

6 0
2 years ago
20 POINTS Use the problem below to answer the question: 34 grams of carbon react with an unlimited amount of H2O. The reaction i
Simora [160]

The carbon would be your answer hope this helps

4 0
2 years ago
Vamos para playa oh te rajas lol
Anuta_ua [19.1K]

Answer:

let's go to the beach or u will relax lol

6 0
2 years ago
Equal masses of water undergo condensation, deposition, evaporation, and sublimation. a) Which process releases the most energy?
Artemon [7]

Answer:

A. Condensation

B. Evaporation

Explanation:

Condensation releases energy when water vapor condenses to form water droplets. Evaporation absorbs energy whenever it changes from liquid to gas, the heat from the sun heats the water up and absorbs energy.

6 0
2 years ago
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0
3 years ago
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