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3 years ago

Which metal will more easily lose an electron sodium or potassium?

1 answer:

7 0

The metal that would more easily lose an electron would be potassium. It is more reactive than sodium. Also, looking on the periodic table, from top to bottom for groups 1 and 2, reactivity increases. So, it should be potassium. Hope this answers the question. Have a nice day.

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In a reaction 4 NH3 + 5 O2 → 4 NO + 6 H2O, 1 mole of ammonia reacts with 2 moles of oxygen. Which of these is correct after the

Vsevolod [243]

Answer:

Option B) All the ammonia is consumed

Explanation:

1. Balanced chemical equation (given):

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

2. Theoretical mole ratios:

$4molNH_3:5molO_2:4molNO:6H_2O$

3. Limiting reactant:

When 1 mole of ammonia is combined with 2 moles of oxygen, the mole ratio is:

$1molNH_3:2molO_2$

Hence, one of the reactants will be completely consumed (the limiting reactant) and, after completion, an excess of the other will remain unreacted.

You need to compare the the actual ratio with the theoretical ratio.

4/5 > 1/2

Hence, NH₃ is in less proportion with respect to oxygen than what is theoretically needed, and the former is the limiting reactant.

Therefore, the 1 mole (all) of ammonia will be consumed, while some oxygen will remain as excess. This is described by the option B) All the ammonia is consumed.

4. Analyze the other options:

The amount of oxygen that will react is:

$1molNH_3\times5molO_2/4molNH3=5/4molO_2$

And the amount that will remain is:

$2molO_2-4/5molO_2=6/5molO_2$

Neither option A) nor C) describe that situation.

The amount of water produced is:

$1molNH_3\times 6molH_2O/4molNH_3=1.5molH_2O$ , which is not described by the option D).

Hence, the correct answer is the option B) All the ammonia is consumed.

6 0

3 years ago

Organisms are classified into domains and other groups by similar characteristics. There are three domains: Archaea, Bacteria, a

Anton [14]

Answer:

presence or absence of a nucleus

Explanation:

These classification of organisms into broad domains is based on the present or absence of nucleus in the cell of an organism.

The archaea are prokaryotes and they lack a distinct cellular nuclei.

Bacteria are similar to archaea but bacteria have only one RNA polymerase

Eukarya have true nucleus and membrane bound organelles .

So, the variations in their cells are used to classify organisms into the broad categories

8 0

3 years ago

The ph of 0.015 m hno2 (nitrous acid) aqueous solution was measured to be 2.63. what is the value of pka of nitrous acid?

Licemer1 [7]

Nitrous acid dissociates as follows:

HNO₂(s) ⇄ H⁺(aq) + NO₂⁻(aq)

According to the equation, an acid constant has the following form:

Ka = [H⁺] × [NO₂⁻ ] / [HNO₂]

From pH, we can calculate the concentration of H⁺ and NO₂⁻:

[H⁺] = 10^-pH = 10^-2.63 = 0.00234 M = [NO₂⁻]

Now, the acid constant can be calculated:

Ka = 0.00234 x 0.00234 / 0.015 = 3.66 x 10⁻⁴

And finally,

pKa = -log Ka = 3.44

7 0

3 years ago

When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for

Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

3 0

3 years ago

A sample of an ideal gas has a volume of 2.21 L at 282 K and 1.03 atm. Calculate the pressure when the volume is 1.84 L

PIT_PIT [208]

Answer:

1.33 atm

Explanation:

use general gas equation P1 V1/ T1 = P2 V2/ T2

rearrange and make P2 the subject then solve,it should give you 1.33 atm

3 0

3 years ago