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Oksi-84 [34.3K]
2 years ago
8

What is the total number of moles of H2SO4 needed to prepare 5.0 liters of a 2.0 M solution of H2SO4

Chemistry
2 answers:
Lerok [7]2 years ago
5 0

Answer:

10

Explanation:

forsale [732]2 years ago
3 0
The molarity of H2SO4 is the number of moles in 1 L of solution.
The molarity is 2.0 mol/L
This means that there should be 2 moles in a 1 L solution to make up this molarity.
In this case we need to make up a 5 L solutions with that molarity. Then the amount of moles required are - 2 mol/L x 5 L = 10 mol
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Each substance in a solution mixture is visible to the naked eye? True or False
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False
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4 0
2 years ago
A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
2 years ago
Which element is likely to gain one electron when it combines chemically with another element? a potassium b krypton c gold d ch
11Alexandr11 [23.1K]
Chlorine because it lacks one valence electron to fill it's outer shell
7 0
3 years ago
Consider an electron with charge −e−e and mass mmm orbiting in a circle around a hydrogen nucleus (a single proton) with cha
PtichkaEL [24]

Answer:

Explanation:

The net force on electron is electrostatic force between electron and proton in the nucleus .

Fc = \frac{1}{4\pi\epsilon} \times \frac{e\times e}{r^2}

This provides the centripetal force for the circular path of electron around the nucleus .

Centripetal force required = \frac{m\times v^2}{r}

So

\frac{m\times v^2}{r}=\frac{1}{4\pi\epsilon} \times \frac{e\times e}{r^2}

v^2=\frac{e^2}{4\pi \epsilon m r}

v=(\frac{e^2}{4\pi \epsilon m r})^{\frac{1}{2} }

5 0
2 years ago
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tensa zangetsu [6.8K]

Answer:

atomic mass of X is 48.0 amu

Explanation:

Let y be the atomic mass of X

Molar mass of O_2 is = 2×16 = 32 g / mol

X + O2 -----> XO_2

According to the equation ,

y g of X reacts with 32 g of O_2

24 g of X reacts with Z g of O_2

Z = ( 32×24) / y

But given that 24.0 g of X exactly reacts with 16.0 g of O_2

So Z = 16.0

⇒ (32×24) / y = 16.0

⇒ y = (32×24) / 16

y= 48.0

So atomic mass of X is 48.0 amu

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