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Alex787 [66]
2 years ago
8

PLZ HELP ME WITH MY WORK​

Chemistry
2 answers:
viktelen [127]2 years ago
8 0
Mixture








I’m sure mann
IRISSAK [1]2 years ago
3 0

Answer:

Explanation:

pretty sure its mixure.

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What are the pH of these solutions?
Tju [1.3M]

Answer:

The answer to your question is below

Explanation:

a)    HCl  0.01 M

      pH = -log [0.01]

      pH = - (-2)

     pH = 2

b)    HCl = 0.001 M

      pH = -log[0.001]

      pH = -(-3)

      pH = 3

c)    HCl = 0.00001 M

       pH = -log[0.00001]

       pH = - (-5)

      pH = 5

d) Distilled water

      pH = 7.0

e) NaOH = 0.00001 M

       pOH = -log [0.00001]

       pOH = -(-5)

       pH = 14 - 5

       pH = 9

f)  NaOH = 0.001 M

      pOH =- log [0.001]

      pOH = 3

      pH = 14 - 3

      pH = 11

g)   NaOH = 0.1 M

       pOH = -log[0.1]

       pOH = 1

       pH = 14 - 1

       pH = 13

3 0
2 years ago
Explain why beryllium is produced when potassium is heated with beryllium
Vladimir [108]

Explanation:

The more reactive element replaces less reactive element during chemical reaction.

Since, potassium is more reactive than beryllium. When potassium reacts with beryllium choride, it replaces beryllium and forms potassium chloride and produces beryllium.

3 0
2 years ago
A boat covers 2,000 meters in 30 minutes. How fast are they traveling
brilliants [131]
1+1 is 2 hope this helps
3 0
3 years ago
Silver (Ag) has two stable isotopes: 107Ag, 106.90 amu, and 109Ag, 108.90 amu. If the average atomic mass of silver is 107.87 am
inysia [295]

Answer:

  • The abundance of 107Ag is 51.5%.
  • The abundance of 109Ag is 48.5%.

Explanation:

The <em>average atomic mass</em> of silver can be expressed as:

107.87 = 106.90 * A1 + 108.90 * A2

Where A1 is the abundance of 107Ag and A2 of 109Ag.

Assuming those two isotopes are the only one stables, we can use the equation:

A1 + A2 = 1.0

So now we have a system of two equations with two unknowns, and what's left is algebra.

First we<u> use the second equation to express A1 in terms of A2</u>:

A1 = 1.0 - A2

We <u>replace A1 in the first equation</u>:

107.87 = 106.90 * A1 + 108.90 * A2

107.87 = 106.90 * (1.0-A2) + 108.90 * A2

107.87 = 106.90 - 106.90*A2 + 108.90*A2

107.87 = 106.90 + 2*A2

2*A2 = 0.97

A2 = 0.485

So the abundance of 109Ag is (0.485*100%) 48.5%.

We <u>use the value of A2 to calculate A1 in the second equation</u>:

A1 + A2 = 1.0

A1 + 0.485 = 1.0

A1 = 0.515

So the abundance of 107Ag is 51.5%.

3 0
2 years ago
he number-average molecular weight of a polypropylene is 1,000,000 g/mol. Compute the degree of polymerization.
m_a_m_a [10]

Answer:

The answer is "23765.4"

Explanation:

Motor weight average number (\bar{M_n}) = 1000000 \frac{g}{mol}

Poly condensation degree dependent on the average number of molecular weights is as follows:

DP_n = \frac{\text{Mol.Wt Number Medium}}{\text{Monomer Unit Mol.Wt}}

All monomer module, in this case, is propylene  

Sunrise. Unit Wt = Mol. Propylene weight

                           = \ Mol. \ Wt \ of \ C_3H_6\\\\= 3 \times 12.01 +6 \times 1.008 \ \ \frac{g}{mol}\\\\= 42.078 \frac{g}{mol}\\\\

DP_n = \frac{1000000}{42.078}\\\\

    =23765.4

8 0
2 years ago
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