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3 years ago

Which organisms can reproduce through budding?

1 answer:

7 0

It's sponges because they reproduce through budding. Ants have a queen and so do bees. And Strawberries reproduced through seeds and flowers

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The equilibrium constant Kc for the reaction below is 0.00584 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentra

Vinil7 [7]

Answer:

Explanation:

Given that:

The chemical equation for the reaction is:

Br2(g) ⇌ 2Br(g)

Initially 0.0345M 0.0416M

$Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016$

$Q_C =0.05016 >>> K_c(0.00584)$

Thus, the given reaction will proceed in the backward direction

The I.C.E table is as follows:

Br2(g) ⇌ 2Br(g)

I 0.0345 0.0416

C +x -2x

E (0.0345+x) (0.0416 -2x)

$K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584$

= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)

= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x

= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908 - 0.17184x + 4x²

Solving by using Quadratic formula

x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] = (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] = (0.0416 -2(0.0126)) = 0.0164 M

4 0

3 years ago

Value of MA is always lesser thàn 1 why?

Degger [83]

Answer:

because the load arm is greater than the effort arm. As we know that when the load arm is greater than the effort arm, the mechanical advantage will always be lesser than one, which results in gain in speed

3 0

2 years ago

Is frying an egg a physical or chemical change

Lemur [1.5K]

Chemical change. The heat from frying an egg causes the egg's protein to change permanently in texture and in appearance through a process.

7 0

3 years ago

A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th

kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = $\frac{38.8}{61.2^{2} }$

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

0.0104 = $\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}$

0.0104 $[P(NO_{2} )]^{2}$ + $P(NO_{2} )$ - 200 = 0

Resolving the second degree equation:

$P(NO_{2} )$ = $\frac{-1+\sqrt{9.32} }{0.0208}$

$P(NO_{2} )$ = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are $P(NO_{2} )$ = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0

3 years ago

Differences between a weak acid and a dilute acid

nika2105 [10]

A weak acid has a low concentration of H+ Ions and a dilute acid is a solution where acid is dissolved in a more volume of water than that of acid.

4 0

3 years ago

Read 2 more answers