Which of these compounds would most likely be found in a deposit of natural gas?

The heating curve below shows the temperature change that occurs as a solid is heated. What is occurring at segment BC of the gr

Anettt [7]

If Bc is the first horizontal line there, the solid is melting and going into the liquid state

6 0

3 years ago

What Volume Of Oxygen react with carbon monooxide to produce 0.5 mole

diamong [38]

Answer:

6000cm3

Explanation:

2CO+O2-------> 2CO2

CO:CO2

1: 1

CO= 0.5 moles

CO:O2

2:1

O2 moles= 0.5÷2

=0.25moles

1 mole------> 24000cm3

0.25moles

(0.25x24000)÷1

= 6000cm3

4 0

4 years ago

Help me with number 4

nalin [4]

The 4th one I’m pretty sure.

6 0

3 years ago

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the

Nataliya [291]

Answer: The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.

Explanation:

Given: Concentration of hydrogen fluoride = 0.126 M

Concentration of fluoride ions = 0.1 M

Volume of HCl = 9.0 mL

Concentration of HCl = 0.01 M

Volume of HCl = 25.0 mL

Moles of $F^{-}$ ions are calculated as follows.

$Moles of F^{-} = molarity \times volume\\= 0.1 M \times 0.025 L\\= 0.0025 mol$

Moles of HF are as follows.

$Moles of HF = Molarity \times Volume\\= 0.126 M \times 0.025 L\\= 0.00315 mol$

Moles of HCl are as follows.

$Moles of HCl = Molarity \times volume\\= 0.01 M \times 0.009 L\\= 0.00009 mol$

Now, reaction equation with initial and final moles will be as follows.

$H^{+} + F^{-} \rightarrow HF$

Initial: 0.00009 0.0025 0.00315

Equilibrium: (0.0025 - 0.00009) (0.00315 + 0.00009)

= 0.00241 = 0.00324

Total volume = (9.00 mL + 25.0 mL) = 34.0 mL = 0.034 L

Hence, concentration of fluoride ions is calculated as follows.

$Concentration = \frac{moles}{volume}\\= \frac{0.00241 mol}{0.034 L}\\= 0.0709 M$

Thus, we can conclude that concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.

7 0

3 years ago

The rate constant of a particular first order reaction is 5.45 x 10^-2 sec^-1 at 40.0 oC. What is the rate constant of this reac

marishachu [46]

<u>Answer:</u> The rate constant for the reaction at 65°C is $0.350s^{-1}$

<u>Explanation:</u>

To calculate rate constant at 65°C of the reaction, we use Arrhenius equation, which is:

$\ln(\frac{K_{65^oC}}{K_{40^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{65^oC}$ = equilibrium constant at 65°C = ?

$K_{40^oC}$ = equilibrium constant at 40°C = $5.45\times 10^{-2}s^{-1}$

$E_a$ = Activation energy of the reaction = 65.5 kJ/mol = 65500 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $40^oC=[40+273]K=313K$

$T_2$ = final temperature = $65^oC=[65+273]K=338K$

Putting values in above equation, we get:

$\ln(\frac{K_{65^oC}}{5.45\times 10^{-2}})=\frac{65500J/mol}{8.314J/mol.K}[\frac{1}{313}-\frac{1}{338}]\\\\K_{65^oC}=0.350s^{-1}$

Hence, the rate constant for the reaction at 65°C is $0.350s^{-1}$

5 0

4 years ago