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VLD [36.1K]
2 years ago
11

A. What is the pH of 8.9 × 10−3 M HCl? B. What is the pH of 8 × 10−8 M HCl?

Chemistry
1 answer:
Leya [2.2K]2 years ago
7 0

Answer:

A. p(H) = 2.0506

B. p(H) = 7.0969

Explanation:

A.

p(H)= -log[H+]

p(H)= -log(8.9×10^-3)

p(H) = 2.0506

B.

p(H)= -log[H+]

p(H)= -log(8×10^-8)

p(H) = 7.0969

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What is the force of gravity acting on an object? (3 points)
oee [108]

the answer of these question is weight

7 0
2 years ago
Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Cu2++2e−→Cu;E∘=0
Maru [420]

Answer:

Explanation:

To calculate the cell potential we use the relation:

Eº cell = Eº oxidation + Eº reduction

Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative,  the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity  because the only thing we need to do is change the sign of the reduction potential for the oxized species .

So the species that is going to be oxidized is the Aluminium, and therefore:

Eº cell = -( -1.66 V ) + 0.340 V =  5.06 V

Equally valid is to write the equation as:

Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species

These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.

3 0
3 years ago
What volume of Co2 (carbon (iv) oxide)
hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

7 0
3 years ago
The chemical formula for table sugar is C 12 H 11 O 22 . What can you tell from this formula?
Paraphin [41]
The the last one, but isn't it c6h12o6?
8 0
3 years ago
4Fe + 3O2 → 2Fe2O3
Black_prince [1.1K]

n(2Fe2O3)=10g/319.374amu=0.03mol

n(4Fe+3O2)=0.03 mol

m(4Fe+2O2)=Mn=319.374×0.03=9.58=10

7 0
3 years ago
Read 2 more answers
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