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3 years ago

In a similar way to oxidation-reduction reactions, half-reactions can be written for electrochemical reactions.

Chemistry

2 answers:

kirza4 [7]3 years ago

7 0

the answer to this question would be true just thought u should know lol xd

irinina [24]3 years ago

7 0

Answer: True

Explanation:

Since in electrochemical reactions, there should be two half reactions called oxidation half cell and Reduction half cell.

Example , reaction between ZN and Cu+2

Oxidation half cell. : Zn----> Zn+2 + 2e-

Reduction half cell: Cu+2 +2e- -----> Cu

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If the solubility of sugar at 25ºC is 211 g/100 g H2O, what mass of sugar can be dissolved in 300 g of H2O?

ladessa [460]

Answer:

633 grams of sugar can be dissolved in 300 g of H₂O

Explanation:

Solubility is the measure of the ability of a certain substance to dissolve in another and form a homogeneous system. Solubility is then the maximum amount of a solute that a solvent can receive and is expressed by concentration units.

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

Then:

$x=\frac{c*b}{a}$

You can apply the rule of three as follows: if by definition of solubility in 100 grams of H₂O there are 211 grams of sugar, in 300 g of H₂O how much sugar is there?

$sugar=\frac{300 grams of H_{2}O *211 grams of sugar}{100 grams of H_{2}O}$

sugar= 633 grams

633 grams of sugar can be dissolved in 300 g of H₂O

7 0

2 years ago

Read 2 more answers

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

Answer: The limiting reactant is ammonia $(NH_3)$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For ammonia:

Given mass of ammonia = 135.9 kg = 135900 g (Conversion factor: 1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol$

For carbon dioxide gas:

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol$

The given chemical reaction follows:

$2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)$

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = $\frac{1}{2}\times 7994.12=3997.06mol$ of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia $(NH_3)$

5 0

3 years ago

Pls Help!!!!! ASAP!!!!

HACTEHA [7]

Answer:

do u need help mentally?

Explanation:

7 0

2 years ago

PLEASE HELP ME!

harina [27]

The pH scale is used to measure the degree of acidity or alkalinity of a solution. The scale runs from 0 (very acidic solutions can have a negative pH) to 14 (very alkaline solutions can have a pH higher than this), while a neutral liquid such as pure water has a pH of 7. The pH is linked to the concentration of hydrogen ions (H +) in the solution. Diluting an acid or alkali affects the concentration of H + ions in a solution and therefore affects the pH. In this activity, we will investigate how diluting an acid or alkali affects the pH.
Hope this helps:D
Have a great rest of a brainly day!

8 0

3 years ago

A solution was found to have a 15.6 % transmittance at 500 nm, its wavelength of maximum absorption, using a cell with a path le

Gnesinka [82]

For the absorbance of the solution in a 1.00 cm cell at 500 nm is mathematically given as

A’ = 0.16138

<h3>What is the absorbance of the solution in a 1.00 cm cell at 500 nm?</h3>

Absorbance (A) 2 – log (%T) = 2 – log (15.6) = 0.8069

Generally, the equation for the Beer’s law is mathematically given as

A = ε*c*l

0.8069 = ε*c*(5.00 )

ε*c = 0.16138 cm-1

then for when ε*c is constant

l’ = 1.00

A’ = (0.16138 cm-1)*(1.00 cm)

A’ = 0.16138

In conclusion, the absorbance of the solution in a 1.00 cm cell at 500 nm is

A’ = 0.16138