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3 years ago

How many electrons are needed for 1 coulomb of negative charge?

Chemistry

1 answer:

Nadusha1986 [10]3 years ago

8 0

Answer:

Explanation:

A single electron has a charge of 1.60217733 × 10-19 Coulombs. A collection of 6.2415 × 1018 electrons has a charge of one Coulomb (1/1.60217733x10-19). 1.6 into 10 raise to power -19 no of electrons are present in one coulomb of charge.

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Determina el pH del café cuya concentración de H+ es de 0.00001M

lara31 [8.8K]

Answer:

Explanation:

Given parameters:

Hydrogen ion concentration = 0.00001M

Unknown:

pH of the solution =?

Solution:

The pH is used to estimate the degree of acidity or alkalinity of a solution. To solve for pH of any solution, we use the expression below;

pH = -log [H⁺]

[H⁺] is the hydrogen ion concentration

pH = -log (1 x 10⁻⁵)

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3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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How many electrons are needed for 1 coulomb of negative charge?​

How many electrons are needed for 1 coulomb of negative charge?