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Jet001 [13]
3 years ago
6

For the reaction below, if the rate of appearance of Br2 is 0.180 M/s, what is the rate of disappearance of

Chemistry
1 answer:
brilliants [131]3 years ago
7 0

Answer:

–0.360 M/s

Explanation:

You might be interested in
What is the number of moles of solute in 250 mL of a 0.4 M solution?
mafiozo [28]

Answer:

0,1 mol

Explanation:

We know that the formula of concentration is C= moles of solute/ volume  

0,4 M= moles of solute/ 250 mL

Convert mL to L      250 mL =0,25 L

0,4 M x 0,25 L= moles of solute

0,1 moles= moles of solute

3 0
3 years ago
10. If 3.5 kJ of energy are added to a 28.2 g sample of iron at 20°C, what
igor_vitrenko [27]

Answer:

569K

Explanation:

Q = 3.5kJ = 3500J

mass = 28.2g

∅1 = 20°C = 20 + 273 = 293K

∅2 = x

c = 0.449

Q = mc∆∅

3500 = 28.2×0.449×∆∅

3500 = 12.6618×∆∅

∆∅ = 3500/12.6618

∆∅ = 276.4220

∅2 - ∅1 = 276.4220

∅2 = 276.4220 + ∅1

∅2 = 276.4220 + 293

∅2 = 569.4220K

∅2 = 569K

3 0
2 years ago
For the reaction: 3 H2(g) + N2(g) <--> 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [
masya89 [10]

The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.

Read more on equilibrium constant;

brainly.com/question/1619133

#SPJ1

5 0
2 years ago
A solution is made by combining 500 mL of 0.10 M HF (Ka=7.2 x 10^-4) with 300 mL of 0.15 M NaF. What is the pH of the resulting
n200080 [17]

Answer:

b) 3.10

Explanation:

HF ⇄ H + + F

Using Henderson-Hasselbalch Equation:

pH = pKa + log [A-]/[HA].

Where;

pKa = Dissociation constant = -log Ka

Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266

[A-] = concentration of conjugate base after dissociation = moles of base/total volume

          = 0.15 x 0.3/0.8

               = 0.05625 M

[HA] = concentration of the acid = moles of acid/total volume

             = 0.10 x 0.5/0.8

                    = 0.0625 M

Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>

pH = 3.14266 + log [0.05625/0.0625]

      = 3.14267 + (-0.04575749056)

           = 3.09691250944

<em>From all the available options below:</em>

<em>a) 2.97 </em>

<em>b) 3.10 </em>

<em>c) 3.19 </em>

<em>d) 3.22 </em>

<em>e) 3.32</em>

The correct option is b.

4 0
3 years ago
Help with this question I got so confused 30 points
frutty [35]

Answer:

B:the second answer

Explanation:

Because the all have leaves

6 0
3 years ago
Read 2 more answers
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