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ICE Princess25 [194]

3 years ago

13

What is the potential energy of a 2-kg vase sitting on a table with a height of 1 meter?

1 answer:

lisov135 [29]3 years ago

7 0

"20 J" is the "Potential Energy" of "vase" on the table.

<u>Explanation</u>:

The body possess some energy according to the position of the object that is the when an object is placed at a height it stores some energy it is called "Potential Energy".

Formula used to calculate the "Potential Energy" is $m\times g \times h$

Where, "m" is mass of vase is given that 2 kg, "g" is acceleration due to gravity is $10 m/s^2$ and "h" is height of the object where it placed is 1 m.

Substitute the given values in the formula,

Potential energy of the vase = $2\times10\times1$

Potential energy of the vase = 20 J

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Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec

Svetach [21]

The overall reaction is given by:

$Br_{2}(g) + 2 NO(g) \rightarrow 2 NOBr(g)$

The fast step reaction is given as:

$NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})$

The slow step reaction is given as:

$NOBr_{2}(g) + NO(g) \rightarrow 2 NOBr(g)$ (slow step $k_{2}$ )

Now, the expression for the rate of reaction of fast reaction is:

$r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]$

The expression for the rate of reaction of slow reaction is:

$r_{2}=k_{2}[NOBr_{2}] [NO]$

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as $[NOBr_{2}]$ takes place in this reaction.

The expression of rate of formation is:

$\frac{d(NOBr)}{dt}=r_{2}$

= $k_{2}[NOBr_{2}][NO]$ (1)

Now, consider that the fast step is always is in equilibrium. Therefore, $r_{1}=0$

$k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]$

$[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]$

Substitute the value of $[NOBr_{2}]$ in equation (1), we get:

$\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]$

= $k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]$

= $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$

Thus, rate law of formation of $NOBr$ in terms of reactants is given by $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$ .

4 0

3 years ago

PLEASE HELP!!!! 20 MINS LEFT IM DYING PLS HELP

Zepler [3.9K]

Answer:

Kc = 0.20

Explanation:

N₂O₄ ⇄ 2NO₂

moles 5.3mol 2.3mol

Vol 5L 5L

Molarity 5.3/5M 2.3/5M

= 1.06M = 0.46M

Kc = [NO₂]²/[N₂O₄] = (0.46)²/(1.06) = 0.1996 ≅ 0.20

6 0

3 years ago

What method is used for chaff from grain<br>

Marta_Voda [28]

Answer:

Winnowing

Explanation:

Wind blows the lighter(in terms of mass) chaff from the whole grains,which are heavier(in terms of mass)

4 0

2 years ago

In the covalent bond between hydrogen and chlorine in the HCl molecule, where will the majority of shared electrons be found?

Mrac [35]

Nearer to the chlorine as it has a greater electronegativity

5 0

3 years ago

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The disaccharide lactose can be decomposed into its constituent sugars galactose and glucose. This decomposition can be accompli

n200080 [17]

Answer:

Rate = 116m⁻¹s⁻¹[lactose][H]⁺

Explanation:

the formula for rate of reaction is given as

Rate = k[lactose]∧α[H]⁺∧β

we solve for the value of α and β

([lactose]₁/[lactose]₂)∧α

α = $\frac{ln\frac{0.00116}{0.00232} }{ln\frac{0.01}{0.02} }$

when we divide this equation

α = $\frac{ln0.5}{ln0.5}$

α = 1

we find β

R₁/R₂ = 0.01/0.02(0.001/0.001)∧β

0.00116/0.00232 = 0.5(1)∧β

β = 1

Rate = k[lactose]∧α[H]⁺∧β

we have to find the value for k

k = 0.00116/0.01(0.001)

k = 0.00116/0.00001

= 116m⁻¹s⁻¹

<u>Rate = 116m⁻¹s⁻¹[lactose][H]⁺</u>

3 0

3 years ago