1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Scrat [10]
3 years ago
9

Consider the following reaction:

Chemistry
1 answer:
Katyanochek1 [597]3 years ago
3 0

Answer:

The three statements are true

Explanation:

For the reaction:

I₂O₅(s) + 5CO(g) → I₂(s) + 5CO₂(g)

State oxidation of iodine in I₂O₅ is:

5 O²⁻ = 10⁻

As you have 2 I and the molecule has no charge, <em>oxidation state of I is +5</em>.

The carbon in CO has an oxidation state of +2 and in CO₂ is +4. That means <em>the carbon is oxidized</em>

<em />

An oxidizing agent is a substance that produce the oxidation of the agent that reacts with this one. CO is oxidized because of I₂O₅ is producing its oxidation being <em>the oxidizing agent</em>

<em></em>

Thus,<em> the three statements are true</em>.

You might be interested in
1. What is a taste bud?
tankabanditka [31]

O tissues containing nerve cells for taste

8 0
2 years ago
Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result
Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0
2 years ago
Fill in the blank Cells break down sugar and release ___________ into the plant.
katrin2010 [14]

Answer:

oxygen

Explanation:

8 0
2 years ago
12. What is the frequency of a photon with an energy of 3.03 x 10-19 J?
bazaltina [42]

Answer:

u=4.57x10^5GHz

Explanation:

Hello.

In this case, given the formula:

E=h*u

Whereas E is the energy, h the Planck's constant and u the frequency of the photon. Thus, solving for it, we obtain:

u=\frac{E}{h}=\frac{3.03x10^{-19}J}{6.63x10^{-34}J*s}\\  \\u=4.57x10^{14}s^{-1}

Or also:

u=4.57x10^{14}Hz*\frac{1GHz}{1x10^9Hz}\\ \\u=4.57x10^5GHz

Best regards.

5 0
3 years ago
Which statement describes how NO2- reacts in this equilibrium:
WITCHER [35]
The statement which describes how NO2- reacts in this equilibrium:
<span>H2SO3(aq) + NO2-(aq) HSO3-(aq) + HNO2(aq
is the second option - </span><span>B. as a Brønsted-Lowry base by accepting a proton.
</span>This is because bases take proton H+ in order to become HNO2. 
6 0
3 years ago
Read 2 more answers
Other questions:
  • Calculate the mass of glucose required to prepare 100ml of 60.0 mM glucose solution .
    15·1 answer
  • What is the only type of invertebrate that may be smarter than some vertebrates?
    10·2 answers
  • A force of 24 N acts on an 8 kg rock. What is the acceleration of the rock?
    14·1 answer
  • During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of
    14·1 answer
  • Why does the pH of the acid solution initially increase very slowly when metal is first added to the acid solution, but graduall
    10·1 answer
  • Why did Chief John Ross decided to sign a treaty with Albert Pike? A. The victory at Wilson’s Creek convinced Ross the Confedera
    11·2 answers
  • Atoms of which elements form bonds without satisfying the octet rule?
    12·1 answer
  • For the reaction between ammonium phosphate and lead (IV) nitrate, producing ammonium nitrate and lead (IV) phosphate, how many
    14·1 answer
  • Given the nuclear equation below, this equation is an example of
    15·1 answer
  • What is the volume of 58.0g of chlorine gas at stp
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!