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IRINA_888 [86]
3 years ago
9

Acid formulas can be distinguished from base formulas because acid formulas usually and base formulas .

Chemistry
1 answer:
sineoko [7]3 years ago
3 0

Answer:

A.  begin with hydrogen; pair a metal with a hydroxide ion

Explanation:

Acid can be distinguished from base on the basis of formulas in such a way that acid formula start with hydrogen.

For example:

Hydrochloric acid is strong acid. Its formula is HCl.

It start with hydrogen

Sulfuric acid is acid. Its formula is H₂SO₄

it is also start with hydrogen.

Base:

Base is distinguish from acid in such a way that in base formula metal is pair with hydroxide ion.

For example:

Sodium hydroxide is base. Its formula is NaOH.

sodium is metal while OH is hydroxide ion.

Potassium hydroxide is base. its formula is KOH.

potassium is metal and OH is hydroxide ion.

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126,720 inches are in 2.0 miles
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The movement of charged particles cannot pass through an electrolyte to produce an electric current.
Rus_ich [418]

The movement of charged particles cannot pass through an electrolyte to produce an electric current.


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FALSE

5 0
3 years ago
A chemistry teacher is able to grade chemistry labs at a rate of 5 labs for every 3 minutes. They need to grade 165 labs how man
PIT_PIT [208]

Answer:

1.65hr

Explanation:

Given parameters:

Number of labs  = 5 labs

 Time taken  = 3 minutes

Unknown:

Time taken in hours to grade 165 labs  = ?

Solution:

Let us find the rate of the teacher;

  Rate  = \frac{number of labs}{time}  

  Insert the parameters and solve;

   Rate  = \frac{5}{3}   = 1.67labs/min

Now;

   Time to grade 165 labs  = \frac{number of labs}{rate}  

   Time to grade 165 labs  = \frac{165}{1.67}   = 98.8min

  Since;

               60min = 1hr

               98.8min  = \frac{98.8}{60}   = 1.65hr

3 0
3 years ago
What element has 2 electrons in the 4s sub level?
ruslelena [56]

is the Calcium because the calcium in group two which group two has 2 electron and also calcium and Potassium 4s sub level group potassium has 1 electron. :)


4 0
3 years ago
Calculate the concentration of all species in a 0.160 m solution of h2co3.
enot [183]

             H₂CO₃ ⇔       HCO₃⁻   +     H⁺

I            0.160               0                 0

C            -x                  +x               +x

E          0.160-x          +x                +x

Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]

4.3 x 10⁻⁷ = x² / (0.160-x)   (x is neglected in 0.160-x = 0.160)

x² = 6.88 x 10⁻⁸

x = 2.62 x 10⁻⁴

             HCO₃⁻    ⇔            CO₃⁻²    +   H⁺

I          2.62 x 10⁻⁴               0                2.62 x 10⁻⁴

C          -x                            +x              +x

E       2.62 x 10⁻⁴ - x           +x              2.62 x 10⁻⁴ + x

Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]

5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)

x = 5.6 x 10⁻¹¹

Thus,

[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M

[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M

[CO₃⁻²] = 5.6 x 10⁻¹¹ M

[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M

[OH⁻] = 3.8 x 10⁻¹¹

8 0
3 years ago
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