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3 years ago

In the Periodic Table, what is the relationship between two elements that are placed side-by-side in a row?

1 answer:

5 0

Answer is D. The periodic table rows are arranged by increasing atomic number. The atomic number is decided by how many protons they have.

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After standardizing a NaOH solution, you use it to titrate an HCl solution known to have a concentration of 0.203 M. You perform

jek_recluse [69]

Answer:

0.203 is the mean of the concentration of the HCl solution

Explanation:

You have 5 concentrations. The most appropiate result is the mean of those results. The mean is a statistical defined as the sum of each result divided by the total amount of results. For the results of the problem, the mean is:

0.210 + 0.204 + 0.201 + 0.202 + 0.197 = 1.014 / 5 =

<h3>0.203 is the mean of the concentration of the HCl solution</h3>

8 0

2 years ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

Finally, by removing the oxidation states we have:

$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0

3 years ago

How would planting a variety of plants in a vacant (empty) lot help establish an ecosystem in that location?

vichka [17]

the plants would grow and reproduce working together forming nutrients from their dropped leaves / branches etc causing insects to come along and do the same along with animals and a keystone species to form a revolving ecosystem continuing an energy moving process

3 0

3 years ago

The condition of acidosis can also cause ____________ because the higher H concentration diffuses to the ICF, pushing K towards

Rina8888 [55]

Answer:

Hyperkalemia

Explanation:

The condition of acidosis can also cause Hyperkalemia because the higher H concentration diffuses to the ICF, pushing K towards the ECF.

4 0

2 years ago

. What mass of ammonium chloride must be added to 250. mL of water to give a solution with pH

Eduardwww [97]

The mass of ammonium chloride that must be added is : ( A ) 4.7 g

<u>Given data :</u>

Volume of water ( V ) = 250 mL = 0.25 L

pH of solution = 4.85

Kb = 1.8 * 10⁻⁵

Kw = 10⁻¹⁴

Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :

NH₄CI + H₂O ⇄ NH₃ + H₃O⁺

where conc of H₃O⁺

[ H₃O⁺ ] = $\sqrt{Ka.C}$ and Ka = Kw / Kb

∴ Ka = 5.56 * 10⁻¹⁰

Next step : Determine the concentration of H₃O⁺ in the solution

pH = - log [ H₃O⁺ ] = 4.85

∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵

Next step : Determine the concentration of NH₄CI in the solution

C = [ H₃O⁺ ]² / Ka

= ( 1.14125 * 10⁻⁵ )² / 5.56 * 10⁻¹⁰

= 0.359 mol / L

Determine the number of moles of NH₄CI in the solution

n = C . V

= 0.359 mol / L * 0.25 L = 0.08979 mole

Final step : determine the mass of ammonium chloride that must be added to 250 mL

mass = n * molar mass

= 0.08979 * 53.5 g/mol

= 4.80 g ≈ 4.7 grams

Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g

Learn more about ammonium chloride : brainly.com/question/13050932

8 0

2 years ago