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inna [77]
3 years ago
12

For 30 points! Please help me understand this question

Chemistry
1 answer:
oksian1 [2.3K]3 years ago
7 0

Answer:

D) Adding a catalyst

Explanation:

Adding a catalyst decreases activation energy and allows the reaction to occur more easily.

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Select the example of symbiosis, chosen from the list below that is best described by the statement shown.
ololo11 [35]

Answer:

can you show us the list

Explanation:

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Which one of the following is not equal to 100 meters
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An acid has an acid dissociation constant of 2.8 10–9. What is the base dissociation constant of its conjugate base?
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Read 2 more answers
4. Chemical formula, mg (Cl0₃)2
grin007 [14]

Answer:

0.017mole

0.0033M

Explanation:

Given parameters:

Formula of the compound:

           Mg(ClO₃)₂

Mass of the sample  = 3.24g

Unknown:

Number of moles of the sample = ?

Molarity  = ?

Solution:

The number of moles of any substance is given as:

   Number of moles  = \frac{mass}{volume}  

 Molar mass of Mg(ClO₃)₂  = 24 + 2[35.5 + 3(16)]  = 191g/mol

    Number of moles  = \frac{3.24}{191}   = 0.017mole

Molarity is the number of moles of a solute in a solution:

    Molarity  = \frac{number of moles }{volume}  

 Volume given  = 5.08L

  Molarity  = \frac{0.017}{5.08}  = 0.0033M

5 0
3 years ago
Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c
malfutka [58]

Answer:

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}}  \\

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Best regards.

8 0
3 years ago
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