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algol13
3 years ago
7

How many atoms are in 578g Na

Chemistry
1 answer:
borishaifa [10]3 years ago
5 0

1.51 x 10²⁵atoms

Explanation:

Given parameters:

Mass of Na = 578g

Unknown:

Number of atoms = ?

Solution:

 To find the number of atoms, we must first find the number of moles the given mass contains.

  Number of moles  = \frac{mass}{molar mass}

   molar mass of Na = 23g

 Number of moles =  \frac{578}{23} =  25.13moles

   1 mole of a substance = 6.02 x 10²³atoms

    25.13 mole of Na = 25.13 x  6.02 x 10²³atoms

 This gives  1.51 x 10²⁵atoms of Na

Learn more:

Avogadro's constant brainly.com/question/2746374

#learnwithBrainly

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What is the mass of silver chlorate (191.32 g/mol) that decomposes to release 0.466L of oxygen gas at STP? AgC1036) _AgCl). _026
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Answer : The mass of silver chlorate will be 2.654 grams.

Explanation :

The balanced chemical reaction is,

2AgClO_3\rightarrow 2AgCl+3O_2

First we have to calculate the moles of oxygen gas at STP.

As, 22.4 L volume of oxygen gas present in 1 mole of oxygen gas

So, 0.466 L volume of oxygen gas present in \frac{0.466}{22.4}=0.0208 mole of oxygen gas

Now we have to calculate the moles of silver chlorate.

From the balanced chemical reaction, we conclude that

As, 3 moles of oxygen produced from 2 moles of silver chlorate

So, 0.0208 moles of oxygen produced from \frac{2}{3}\times 0.0208=0.01387 moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

\text{Mass of }AgClO_3=\text{Moles of }AgClO_3\times \text{Molar mass of }AgClO_3

Molar mass of silver chlorate = 191.32 g/mole

\text{Mass of }AgClO_3=0.01387mole\times 191.32g/mole=2.654g

Therefore, the mass of silver chlorate will be 2.654 grams.

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7 0
3 years ago
"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"
Salsk061 [2.6K]

Answer:- The Ka for the acid is 1.53*10^-^5 .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

Ka = [H^+][A^-]\frac{1}{HA}

Where, Ka is the acid ionization constant. Let's plug in the values.

Ka = \frac{X^2}{0.25-X}

Let's calculate the value of X first using the equation:

pH = -log[H^+][/tex]

on taking antilog ob above equation we get:

[H^+]=10^-^p^H

[H^+]=10^-^2^.^7^1

[H^+] = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

Ka=\frac{(0.00195)^2}{0.25-0.00195}

Ka=1.53*10^-^5

So, the value of Ka for butyric acid is 1.53*10^-^5 .

8 0
3 years ago
Read 2 more answers
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