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3 years ago

How much more basic is a solution with pH 10 than a solution with pH 8?

Chemistry

2 answers:

Simora [160]3 years ago

6 0

100 times more back

Basile [38]3 years ago

6 0

I think it is D. 100 times

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Someone said that we can not use sulfric acid to prepare PbSO4 .

Delvig [45]

Answer:

A. we can use sulfuric acid to prepare PbSO4

7 0

3 years ago

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The EPA scientist measures the pH level in one area of a river to be lower than normal and writes that a pollutant must have bee

shutvik [7]

Its an example of inference.

4 0

3 years ago

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Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

Answer: The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

Explanation:

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

Oxidation half reaction: $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

Reduction half reaction: $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

4 0

3 years ago

How many moles of BaSO4 are formed if 0.5 moles of Na2 SO4 react with 60 g of BaCl2? Na2 SO4 + BaCl2 → BaSO4 + 2NaCl

faust18 [17]

the balanced equation for the reaction is as follows
Na₂SO₄ + BaCl₂ ----> 2NaCl + BaSO₄
stoichiometry of Na₂SO₄ to BaCl₂ is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.

number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other

therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl₂ to BaSO₄ is 1:1.

therefore number of BaSO4 moles formed - 0.288 mol

6 0

3 years ago

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EIGHTY POINTS- WILL GET BRAINLIEST NEED ANSWER ASAP

il63 [147K]

ans A. 13 to 3

solution:

Total puppies for sale=6

Total puppies left = 2

hence total puppies sold= 6 - 2 = 4

cost of a puppy= $ 104

so cost of 2 puppies = 104 x 4

=416$

Now for kittens:

Total kittens for sale = 12

Total kittens left = 8

hence total kittens sold=12 - 8 =4

cost of a kitten = $24

so cost of 4 kittens= 24x 4=96$

Now ratio of sales of puppies to kittens = 208$/96$

= 13/3 or 13:3 or 13 to 3

I hope this explanation will help you to understand this problem

5 0

3 years ago