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3 years ago

15

What determines the average kinetic energy of the particles in a gas?

2 answers:

jolli1 [7]3 years ago

7 0

Answer:

D

The temperature.

Explanation:

Bas_tet [7]3 years ago

4 0

The answer is D (the temperature)

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If the volume of a cylinder is 84,300 cubic centimeters then how many millimeters is this equal to

nevsk [136]

Answer:

84,300,000

Explanation:

Just multiply it by 1,000. 1cm = 10mm, and a cube is the length to a sort of third degree, so you take the 10 to its third degree as well and multiply it by 1,000, instead of 10 like you would to find how many millimeters were in an amount of centimeters. If that makes sense.

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3 years ago

Draw the products formed when cholesterol is treated with [1] bh3·thf; [2] h2o2, −oh. indicate the stereochemistry around any st

Andreas93 [3]

Alkene on Hydration yield Alcohols. When Asymmetric Alkenes are treated with water in the presence of acid they follow Markovnikoff rule, and the Hydrogen of incoming reagent goes to that carbon which contain more number of Hydrogen atoms. The reverse (Anti-Markovnikoff) is done by carrying out Hydroboration reaction.In this case the Hydrogen atom of incoming reagent goes to that carbon which contains less number of Hydrogen atoms. In this reaction both -H and -OH adds in syn fashion. The Hydroboration reaction of Chloestrol is shown below,

3 0

3 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

2 years ago

Determina el pH de una solución si la concentración de H es 3.5 x 10 molar, e indicar si es acido o bas

erastovalidia [21]

Answer:

Explanation:

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3 years ago

Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en

Bezzdna [24]

Answer:

10.19 g CO₂

4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:

1 gallon = 3.785 L

1 L = 1000 mL

Now we<u> convert 1.24 gallons to mL</u>:

1.24 gallon * $\frac{3.785L}{1gallon} *\frac{1000mL}{1L} =$ 4693.4 mL

We <u>calculate the mass of Octane</u>:

4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:

CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{8molCO_{2}}{1molOctane}$ * 44 g/mol = 10.19 g CO₂

H₂O ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{9molH_{2}O}{1molOctane}$ * 18 g/mol = 4.69 g H₂O

7 0

3 years ago