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3 years ago

Adding a catalyst to a reaction has an effect similar to

1 answer:

3 0

I will say increasing temperature but you dont have that option on your list, so I would take B. Increasing Concentration.

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A spring-loaded toy gun projects a 6.1 g nerf pellet horizontally. The spring constant is 8.5 N/m, the barrel of the gun is 13 c

gladu [14]

Answer:

0.664 m/s

Explanation:

Conversion to metric unit:

6.1 g = 0.0061 kg

13 cm = 0.13 m

6.6 cm = 0.066 m

As the spring is released from its compression stage, its elastics energy is converted into the pellet's kinetic energy, which is also affected by the work of friction force:

$E_e = E_k + W_f$

$kx^2/2 = mv^2/2 + Fs$

where k = 8.5 N/m is the spring constant, x = 0.066m is the compression length, m = 0.0061 kg is the pellet mass, v is the velocity of the pellet as it leaves the barrel, F = 0.039 N is the constant frictional force, s = 0.13 m is the distance that the friction force applies on the pellet.

$8.5*0.066^2/2 = 0.0061*v^2/2 + 0.039*0.13$

$0.0185 = 0.00305v^2 + 0.00507$

$v^2 = 0.44$

$v = \sqrt{0.44} = 0.664 m/s$

7 0

4 years ago

2H2S(g)⇌2H2(g)+S2(g),Kc=1.67×10−7 at 800∘C is carried out at the same temperature with the following initial concentrations: [H2

scoray [572]

Answer : The concentration of $S_2$ at equilibrium will be, $1.67\times 10^{-7}M$

Explanation : Given,

Equilibrium constant = $1.67\times 10^{-7}$

Initial concentration of $H_2S$ = 0.100 M

Initial concentration of $H_2$ = 0.100 M

Initial concentration of $S_2$ = 0.00 M

The balanced equilibrium reaction is,

$2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

Initial conc. 0.1 0.1 0

At eqm. (0.1-2x) (0.1+2x) x

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

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

$1.67\times 10^{-7}=\frac{(0.1+2x)^2\times (x)}{(0.1-2x)^2}$

By solving the term 'x' by quadratic equation, we get two value of 'x'.

$x=1.67\times 10^{-7}M$

Concentration of $S_2$ at equilibrium = $x=1.67\times 10^{-7}M$

Therefore, the concentration of $S_2$ at equilibrium will be, $1.67\times 10^{-7}M$

6 0

4 years ago

Read 2 more answers

Help plz I don't understand

Vika [28.1K]

#78

Airplane start with initial speed

$v_i = 0$

now the takeoff velocity is given as

$v_f = 300 km/h$

$v_f = 83.33 m/s$

acceelration is given as

$a = 1 m/s^2$

now we have

$v_f - v_i = at$

from above equation we have

$83.33 - 0 = 1(t)$

$t = 83.33 s$

#79

Airplane start with initial speed

$v_i = 0$

now the takeoff velocity is given as

$v_f = 300 km/h$

$v_f = 83.33 m/s$

acceelration is given as

$a = 2 m/s^2$

now we have

$v_f - v_i = at$

from above equation we have

$83.33 - 0 = 2(t)$

$t = 41.7 s$

6 0

4 years ago

your friend asks you for a glass of water and you bring her 5 milimeter of water? is this more or less than what she was probabl

Elan Coil [88]

Less of a glass of water because a glass of water holds approximately 250 mil-liters of water

3 0

4 years ago

Block 1 of mass m1 slides along a frictionless floor and into a one-dimensional elastic collision with stationary block 2 of mas

Elan Coil [88]

<span>Answers: (a) 2.0 m/s (b) 4 m/s

Method:

(a) By conservation of momentum, the velocity of the center of mass is unchanged, i.e., 2.0 m/s.

(b) The velocity of the center of mass = (m1v1+m2v2) / (m1+m2)

Since the second mass is initially at rest, vcom = m1v1 / (m1+m2)

Therefore, the initial v1 = vcom (m1+m2) / m1 = 2.0 m/s x 6 = 12 m/s

Since the second mass is initially at rest, v2f = v1i (2m1 /m1+m2 ) = 12 m/s (2/6) = 4 m/s </span>

7 0

3 years ago