All categories

3 years ago

Cole ate grass and ______while lying after attack.

Chemistry

2 answers:

Delicious77 [7]3 years ago

6 0

Cole ate grass and weed while lying after attack

Daniel [21]3 years ago

5 0

cole ate grass and weed

while laying after attack

You might be interested in

describe the relationship between the age of the rocks and their distance from the mid ocean ridge, please look at the picture t

Artemon [7]

Answer:

I do not know the Answer I'm just trying to get my point

Explanation:

Thank you

5 0

3 years ago

How did gus get ao gay ??

Mumz [18]

Magic, that how

.....

4 0

3 years ago

The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates:

alexira [117]

Answer:

Kc = 6x10⁻⁶

Explanation:

For the reaction:

4NH₃(g) + 3O₂(g) ⇄ 2N₂(g) + 6H₂O(g)

Kc is defined as:

Kc =[N₂]² [H₂O]⁶ / [NH₃]⁴ [O₂]³

The equilibrium concentrations of the gases is -Because volume of the container is 1.00L-:

[N₂] = 2X = 1.96x10⁻³; <em>X = 9.8x10⁻⁴</em>

[H₂O] = 6X; 6ₓ9.8x10⁻⁴ = 5.88x10⁻³

[NH₃] = 0.0150M - 4X = 0.01108M

[O₂] = 0.0150M - 3X = 0.01206M

Replacing in Kc expression:

Kc =[1.96x10⁻³]² [5.88x10⁻³]⁶ / [0.01108M]⁴ [0.01206M]³

8 0

4 years ago

HOW FAST CAN YOU ANSWER, LETS SEE

QveST [7]

Answer:

4 atoms

Explanation:

5 0

3 years ago

Read 2 more answers

7) For the reaction 9A (g) + B (g)  5C(g) + 1/6 D (g), it takes 4 and a half minutes for the concentration of C to increase to

viva [34]

Answer: The correct option is, (C) 0.53

Explanation:

The given chemical reaction is:

$9A(g)+B(g)\rightarrow 5C(g)+\frac{1}{6}D(g)$

The rate of the reaction for disappearance of A and formation of C is given as:

$\text{Rate of disappearance of }A=-\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}$

Or,

$\text{Rate of formation of }C=+\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}$

where,

$\Delta C$ = change in concentration of C = 1.33 M

$\Delta t$ = change in time = 4.5 min

Putting values in above equation, we get:

$\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}=\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}$

$\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{\Delta [C]}{\Delta t}$

$\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{1.33M}{4.5min}$

$\frac{\Delta [A]}{\Delta t}=0.53M/min$

Thus, the decrease in A during this time interval is, 0.53

5 0

3 years ago