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2 years ago

What mass of salt is needed to prepare 600 mL of a 4g/L solution?

Chemistry

1 answer:

OLga [1]2 years ago

5 0

THE ALTERNATIVE IS 4.8g alternative c

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How many moles are in 124.7 g of Ba(OH)2

Alekssandra [29.7K]

Answer:

0.73 mol

Explanation:

No. of moles(n)= Given mass/molar mass.

Given mass=124.7g

Molar mass of Ba(OH)2= Molar mass of (Barium+2Oxygen+2Hydrogen)=137+32+2=171g

No. of moles= 124.7g/171g=0.73 mol

7 0

3 years ago

The (OH) of a given solution =1.00x10-9M. what is this solution s pOH?

german

Answer:

Explanation:

Given parameters:

Concentration of OH⁻ [OH]= 1 x 10⁻⁹M

Solution:

To find the pOH of a solution can be found using the expression below:

pOH = -log₁₀[OH]

[OH] = concentration of the hydroxyl ions

pOH = -log₁₀(1 x 10⁻⁹) = - x -9 = 9

8 0

3 years ago

Read 2 more answers

if you were to compare the mass of the products and reactants in a reaction, you would find that the mass of the products is alw

Salsk061 [2.6K]

If you were to compare the mass of the products and reactants in a reaction, you would find that the mass of the products is <span>equal to the mass of the reactants.</span>

3 0

3 years ago

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

6 0

3 years ago

How many moles of H2SO4 are needed to prepare 200 mL of a 1.0 M solution?

jok3333 [9.3K]

The answer is 0.2moles

4 0

3 years ago