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kiruha [24]
3 years ago
6

Which of the following elements is the most reactive?

Chemistry
2 answers:
Maksim231197 [3]3 years ago
4 0
<span>the following element that is most reactive </span>would be  Fluorine
Marta_Voda [28]3 years ago
4 0
I'm pretty sure its Bromine.
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Baking soda (NaHCO3) can be added to a fruit mix solution to create a carbonated drink. An example is the reaction between bakin
DanielleElmas [232]

Answer:

74.4 ml

Explanation:

          C₆H₈O₇(aq) + 3NaHCO₃(s) => Na₃C₆H₅O₃(aq + 3CO₂(g) + 3H₂O(l)

Given     15g = 15g/84g/mol = 0.1786mole Sodium Bicarbonate

From equation stoichiometry 3moles NaHCO₃ is needed for each mole citric acid or, moles of citric acid needed is 1/3 of moles sodium bicarbonate used.

Therefore, for complete reaction of 0.1786 mole NaHCO₃ one would need 1/3 of 0.1786 mole citric acid or 0.0595 mole H-citrate.

The question is now what volume of 0.8M H-citrate solution would contain 0.0595mole of the H-citrate? This can be determined from the equation defining molarity. That is => Molarity = moles solute / Liters of solution

=> Volume (Liters) = moles citric acid / Molarity of citric acid solution

=> Volume needed in liters = 0.0.0595 mole/0.80M = 0.0744 Liters or 74.4 ml

6 0
3 years ago
In a redox reaction, one of the reactants must release
Leviafan [203]
Redox reaction is the reduction and oxidation reaction. It is a chemical reaction that involves a transfer of electrons. It can mean loss of oxygen (oxidation) or gain of electrons (reduction). Oxydation is the process of where a sustance loses electrons, gains an oxygen atom/s, loses a hydrogen atom/s.  reduction is the opposite.
In a redox reaction, one of the reactants must release electrons.
3 0
3 years ago
You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will t
Nataly [62]
First, we should get moles acetic acid = molarity * volume

                                                                =0.3 M * 0.5 L

                                                                =  0.15 mol

then, we should get moles acetate = molarity * volume

                                                           = 0.2 M * 0.5L

                                                           = 0.1 mol

then, we have to get moles of OH- which added:

moles OH- = molarity  * volume

                   = 1 M    * 0.02L

                  = 0.02 mol

when the reaction equation is:


                 CH3COOH  +  OH-  → CH3COO-   +  H2O


moles acetic acid after adding OH- = (0.15-0.02) 
                                             
                                                            =  0.13M                                       

moles acetate after adding OH- =  (0.1 + 0.02)

                                                      =   0.12 M

Total volume = 0.5 L + 0.02 L= 0.52 L

∴[acetic acid] = moles acetic acid after adding OH- / total volume

                        = 0.13mol / 0.52L

                       = 0.25 M

and [acetate ) = 0.12 mol / 0.52L
 
                        = 0.23 M

by using H-H equation we can get PH:

PH = Pka + ㏒[salt/acid]

when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka 

        = -㏒ 1.8 x 10^-5

       = 4.7

So by substitution:

∴ PH = 4.7 + ㏒[acetate/acetic acid]

         = 4.7 + ㏒(0.23/0.25)

        = 4.66
6 0
3 years ago
Earth's yearly trip AROUND the sun is called what?
dedylja [7]
The answer is revolution
7 0
3 years ago
Read 2 more answers
what is the minimum mass of ethylene glycol that must be dissolved in 14.5 kg of water to prevent the solution from freexing at
lbvjy [14]

Answer:

The minimum mass of ethylene glycol = 6.641 Kg

Explanation:

\Delta T_f= T_f-T_f'\\T_f'=T_f-\Delta T_f

Where T_f = freezing point of pure solvent water, 0°C

T_f'= Freezing point of solvent after mixture

K_f = Freezing point depression constant = 1.86 °C/m

Moecular weight of ethylene glycol = 60 g/mol

Weight of ethylene glycol = 14.5 Kg= 14.5×10^3 g

molality of ethylene glycol

m = \frac{weight}{mol.wt} \times\frac{1000}{V}

Substitute the values to calculate m

m = \frac{w}{60} \times\frac{1000}{14.5\times1000}

by formula

0-(-14.2) =1.86\frac{w}{60} \times\frac{1000}{14.5\times1000}

calculating we get w = 6641.93 g

Therefore, The minimum mass of ethylene glycol = 6.641 Kg

4 0
3 years ago
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