HELP FAST!!!

Help I don't understand balancing equations

Semmy [17]

Answer:

Explanation: so I can help you with this I show you can you some examples that I’ve done in the past years

7 0

1 year ago

At a given temperature, the elementary reaction A − ⇀ ↽ − B , A↽−−⇀B, in the forward direction, is first order in A A with a rat

Svetllana [295]

Answer:

The equilibrium constant for the reversible reaction = 0.0164

Explanation:

At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.

The reaction is given as

A ⇌ B

Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]

The rate of forward reaction = |r₁| = k₁ [A]

The rate of backward reaction = |r₂| = k₂ [B]

(Taking only the magnitudes)

where k₁ and k₂ are the forward and backward rate constants respectively.

k₁ = 0.010 s⁻¹

k₂ = 0.0610 s⁻¹

|r₁| = 0.010 [A]

|r₂| = 0.016 [B]

At equilibrium, the rate of forward and backward reactions are equal

|r₁| = |r₂|

k₁ [A] = k₂ [B] (eqn 1)

Note that equilibrium constant, K, is given as

K = [B]/[A]

So, from eqn 1

k₁ [A] = k₂ [B]

[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164

K = [B]/[A] = (k₁/k₂) = 0.0164

Hope this Helps!!!

5 0

2 years ago

If a 45 mM phosphate solution(solution A) had an absorbance of 1.012. What would be the absorbance if 11 mL of solution A was us

timofeeve [1]

Answer:

0.550

Explanation:

The absorbance (A) of a substance depends on its concentration (c) according to Beer-Lambert law.

A = ε . l . c

where,

ε: absorptivity of the species

l: optical path length

A 45 mM phosphate solution (solution A) had an absorbance of 1.012.

A = ε . l . c

1.012 = ε . l . 45 mM

ε . l = 0.022 mM⁻¹

We can find the concentration of the second solution using the dilution rule.

C₁ . V₁ = C₂ . V₂

45mM . 11mL = C₂ . 20.0 mL

C₂ = 25 mM

The absorbance of the second solution is:

A = (ε . l ). c

A = (0.022 mM⁻¹) . 25 mM = 0.55 (rounding off to 3 significant figures = 0.550)

8 0

2 years ago

Limestone is made of predominantly calcium carbonate. Geologists can test for limestone using hydrochloric acid. The products ar

Blizzard [7]

I'm obsessed with me as much as you
Say you'd die for me I'd die for me too

7 0

2 years ago

Calculate the hight of a column of liquid glycerol in meters required to exert the same pressure as 3.02 m if CCl4?

katovenus [111]

Answer:

Approximately $3.81\; \rm m$ .

Explanation:

Look up the density $\rho$ of carbon tetrachloride, $\rm CCl_4$ , and glycerol:

Density of carbon tetrachloride: approximately $1.59\times 10^{3}\; \rm kg \cdot m^{-3}$ .
Density of glycerol: approximately $1.26\times 10^{3}\; \rm kg \cdot m^{-3}$ .

Let $g$ denote the gravitational field strength. (Typically $g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth.) For a column of liquid with a height of $h$ , if the density of the liquid is $\rho$ , the pressure at the bottom of the column would be:

$P = \rho\cdot g \cdot h$ .

The pressure at the bottom of this carbon tetrachloride column would be:

$\begin{aligned} P &= \rho \cdot g \cdot h \\ & \approx 1.59\times 10^{3} \; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1} \times 3.02 \; \rm m \approx 4.71 \times 10^{4} \; \rm N \cdot m^{-2} \end{aligned}$ .

Rearrange the equation $P = \rho\cdot g \cdot h$ for $h$ :

$\displaystyle h = \frac{P}{\rho \cdot g}$ .

Apply this equation to calculate the height of the liquid glycerol column:

$\begin{aligned}h &= \frac{P}{\rho \cdot g} \\ &\approx \frac{4.71 \times 10^{4}\; \rm N \cdot m^{-2}}{1.26 \times 10^{3}\; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1}} \approx 3.81\; \rm m\end{aligned}$ .

4 0

2 years ago