All categories

4 years ago

Which state of matter expands when heated and is easy to compress

2 answers:

8 0

Hello there,

The answer to your question is:

GAS

Hope this answer helps you :)

P.S. You might find this link useful:

https://quizlet.com/110502712/chemistry-ch2-flash-cards/

Good Luck.

AlladinOne [14]4 years ago

6 0

That would be gases

You might be interested in

How long does it take the Earth to move from Point A to Point C?

e-lub [12.9K]

Answer:

24 hours

Explanation:

8 0

3 years ago

Read 2 more answers

Christy is a forensic scientist. She is given

Nitella [24]

Answer:

The white powder is not sugar, so Christy's hypothesis is incorrect . Christy's next step should be to identify substances that have the measured molecular mass.

Explanation:

6 0

4 years ago

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

What are chemical reaction

scoray [572]

The reaction in which we can't take back the product to reactant form .Or in simple words the reaction which is irreversible is called chemical reaction

Example:-

Burning of wax

We cannot pull back the reaction to get wax back .

3 0

2 years ago

Read 2 more answers

At a given temperature, the elementary reaction A --->B in the forward direction is first order in A with a rate constant of

Pani-rosa [81]

Answer:

The value of the equilibrium constant for the reaction A ⇒ B is Kc = 1.72 × 10³.

The value of the equilibrium constant for the reaction B ⇒ A is K'c = 5.81 × 10⁻⁴.

Explanation:

For the reaction A ⇒ B, the equilibrium constant (Kc) is equal to the forward rate constant (kf) divided by the reverse rate constant (ki).

$Kc=\frac{kf}{ki} =\frac{1.60 \times 10^{2} s^{-1} }{ 9.30 \times 10^{-2} s^{-1}} =1.72 \times 10^{3}$

If we consider the inverse reaction B ⇒ A, its equilibrium constant (K'c) is the inverse of the forward reaction equilibrium constant.

$K'c=\frac{1}{Kc} =\frac{1}{1.72 \times 10^{3} } =5.81 \times 10^{-4}$

4 0

3 years ago