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3 years ago

How much water should be added to 85 mL of 0.45 M HCI to reduce the concentration to 0.20 M?

Chemistry

1 answer:

valkas [14]3 years ago

8 0

Answer:

106.25 mL

Explanation:

For this, we can use

C1×V1=C2×V2

C1 = 0.45

V1 = 85

C2= 0.20

V2= ?

0.45 × 85 = 0.20 × V2

V2= (0.45 × 85)/0.20

V2=191.25mL

To find the amount of water added, subtract V1 from V2

191.25 - 85 =106.25mL

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poizon [28]

We know that each millimeter contains 10⁻³ meters. Writing this as a ratio:
1 mm : 10⁻³ m

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Therefore, the conversion used will be:
(1 mm / 10⁻³ m)³

When we multiply by this conversion, we will get:
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3 years ago

It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta

Zigmanuir [339]

Answer:

$\lambda=241.9\ nm$

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that

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1 mole = $6.023\times 10^{23}\ electrons$

So,

$6.023\times 10^{23}$ electrons can be removed by applying of 495.0 kJ of energy.

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Energy required = $82.18\times 10^{-23}\ kJ$

Also,

1 kJ = 1000 J

So,

Energy required = $82.18\times 10^{-20}\ J$

Also, $E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}$

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3 years ago

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VashaNatasha [74]

There are 18 protons and electrons and 22 neutrons in the atom

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3 years ago

How many grams of KCIO3 are needed to produce 5.00 of O2 at STP?

Arlecino [84]

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3 years ago

A 3.82 g piece of limestone contains 2.62 g of CaCO3

Kobotan [32]

Considering the definition of percentage by mass, the mass percentage of CaCO₃ is 68.59%.

<h3>What is mass percentage</h3>

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