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likoan [24]
3 years ago
12

Convert 15.2 moles of K to atoms of K.

Chemistry
1 answer:
choli [55]3 years ago
3 0

Answer:

\boxed {\boxed {\sf 9.15*10^{24} \ atoms \ K}}

Explanation:

To convert atoms to moles, Avogadro's Number must be used: 6.022*10²³.

This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case it is the atoms of potassium. We can create a ratio.

\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }

Multiply by the given number of moles: 15.2

15.2 \ mol \ K *\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }

The moles of potassium cancel.

15.2 *\frac {6.022*10^{23} \ atoms \ K }{ 1 }

The denominator of 1 can be ignored.

15.2 * {6.022*10^{23} \ atoms \ K }{

Multiply.

9.15344*10^{24} \ atoms \ K

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated that is the hundredth place. The 3 in the thousandth place tells us to leave 5.

9.15*10^{24} \ atoms \ K

In 15.2 moles of potassium, there are <u>9.15*10²⁴ atoms of potassium.</u>

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2 years ago
a 25 ml volume of a sodium hydroxide solution requires 19.6 mL of a 0.189 M HCl acid for neutralization. A 10 mL volume of phosp
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Answer: 0.172 M

Explanation:

a) To calculate theconcentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=0.189M\\V_1=19.6mL\\n_2=1\\M_2=?\\V_2=25mL

Putting values in above equation, we get:

1\times 0.189\times 19.6=1\times M_2\times 25\\\\M_2=0.148

b) To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_3PO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=3\\M_1=?\\V_1=10mL\\n_2=1\\M_2=0.148\\V_2=34.9mL

Putting values in above equation, we get:

3\times M_1\times 10=1\times 0.148\times 34.9\\\\M_1=0.172M

The concentration of the phosphoric acid solution is 1.172 M

3 0
3 years ago
effect. RbClO(s) → Rb+(aq) + ClO−(aq) HClO(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + ClO−(aq) The degree of dissociatio
Lemur [1.5K]

Explanation:

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of strontium sulfate and sodium sulfate follows the equation:

RbClO(s)\rightarrow Rb^{+}(aq.)+ClO^{-}(aq.)

HClO(aq)+H_2O\rightleftharpoons H_3O^+(aq.)+ClO^{-}(aq.)

According to Le-Chateliers principle: If there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, hypochlorite ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of hydrogen hypochlorite.

Thus, the addition hypochlorite ions will shift the equilibrium in the left direction.

The dissociation of hydrogen hypochlorite is suppressed due to the common ion effect.

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3 years ago
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