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Vinvika [58]
3 years ago
14

In chemical reactions, atoms are

Chemistry
1 answer:
Anna007 [38]3 years ago
5 0
<span>In chemical reaction, atoms always are rearranged. They are never destroyed or created because it would contradict the law of conservation of energy. The law of conservation of energy states that a matter cannot be created nor destroyed. It can only be formed in one way or the other. Also it cannot be neutralized. Resulting in one cannot be part of a chemical reaction. An example of this is the reaction HCl + NaOH -> NaCl + H2O. the reactants and products are not created, they are just formed into another substance.</span>
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Rank the nonmetals in each set from most reactive (1) to least reactive (3). bromine chlorine iodine
Anettt [7]

Answer:

The answer is b

Explanation:

have common properties and are good conductors of heat and electricity. They reflect light and are malleable, and ductile.

3 0
3 years ago
Read 2 more answers
How many fluorine atoms bond with calcium to form calcium fluoride? one two three four five
Elena-2011 [213]
Calcium fluoride.
Ca is metal, F is non-metal, so they form ionic bond.
Ca as metal can form only positive ion. Ca in the second group, so the charge of Ca ion is 2+.   Ca²⁺
F is in the 17th group, so it has 7 electrons on the last level. It is non-metal, non-metal, so it has negative charge -(8-7)=-1. "8" because on the last level cannot be more than 8 electrons. F-ion is F¹⁻.

Ca²⁺  F¹⁻
Number of positive charges should be equal to number of negative charges,
Formula of calcium fluoride
CaF2.
2 atoms Fluorine bond with Calcium.
7 0
3 years ago
A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0
3 years ago
A speed time graph shows that a car moves at 20 m/s for 15s. The car's speed steadily decreases until it comes to a stop at 40s.
ladessa [460]
0s to 15s: constant speed/zero acceleration
15s to 40s: constant gradient, therefore constant deceleration
4 0
2 years ago
Calculate e°cell for a silver-aluminum cell in which the cell reaction is al(s) + 3ag+(aq) → al3+(aq) + 3ag(s) –2.46 v 0.86 v –0
tekilochka [14]
When E° cell is an electrochemical cell which comprises of two half cells.
 
So,

when we have the balanced equation of this half cell :

Al3+(aq) + 3e- → Al(s)   and E°1 = -1.66 V 

and we have  also this balanced equation of this half cell :

Ag+(aq)  + e- → Ag(s)  and E°2 = 0.8 V 

so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)

when E° = E°2 - E°1

∴E° =0.8 - (-1.66)

      = 2.46 V

∴ the correct answer is 2.46 V




6 0
3 years ago
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