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2 years ago

how many grams of NaCl will be needed to form 600 ml of a saturated solution at 100 degrees celsius ?

1 answer:

7 0

The solubility of NaCl in water at 100 C is 40%, meaning that we can dissolve 40 g NaCl in 100 g water. Assuming that dissolving NaCl does not add any volume to the solution, 600 mL of water is approximately equal to 600 g of water. By ratio and proportion: 40 g NaCl/100 g H2O = x g NaCl/600 g H2O
x = 240 g NaCl
So 240 g of NaCl must be dissolved to form a saturated solution.

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The black road because during the day it absorbed more radiation than the with one

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2 years ago

Calculate the percent composition for each of the elements in Na3PO4. A 5.00 gram sample of an oxide of lead PbxOy contains 4.33

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Answer:

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Explanation:

The full explanation is found in the attachments below

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2 years ago

For the following equilibrium: A+2B⇋C+3D If the change in concentration for B is 0.44 M, what will be the change in concentratio

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2 years ago

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valkas [14]

Answer:

Explanation:

The given overall reaction is as follows:

O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )

The reaction mechanism for this reaction is as follows:

O ₂ N N H ₂ ⇌ k 1 k − 1 O ₂N N H ⁻ + H ⁺ ( f a s t e q u i l i b r i u m )

O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )

H ⁺ + O H − k ₃→ H ₂ O ( f a s t )

The rate law of the reaction is given as follows:

k = [ O ₂ N N H ₂ ] / [ H ⁺ ]

The rate law can be determined by the slow step of the mechanism.

r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )

Since, from the equilibrium reaction

k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1

[ O ₂ N N H ⁻] = k ₁ /k − 1 × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )

Substitituting the value of equation (2) in equation (1) we get.

r a t e = k ₂ k ₁/ k − 1 × [ O ₂ N N H ₂ ] /[ H ⁺ ]

Therefore, the overall rate constant is

k = k₂k₁/k-1

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2 years ago

A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC

diamong [38]

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

$PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}$

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = $25^oC=273+25=298K$

V = volume = 4.0 L

$n_1$ = moles of $SO_2$ = 0.20 mol

$n_2$ = moles of $O_2$ = 0.20 mol

Now put all the given values in the above expression, we get:

$P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}$

$P=2.4atm$

Thus, the pressure in the flask after reaction complete is, 2.4 atm

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2 years ago