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amid [387]
4 years ago
9

Given the balanced equation:

Chemistry
1 answer:
Marta_Voda [28]4 years ago
7 0
The answer is (2) A bond is formed and energy is released. The left side of equation is I atom and the right side of equation is I2 molecule. So the bond is formed between I atom to form I2 molecule. And forming bond will release energy while breaking bond will absorb energy.
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At a particular temperature, an equilibrium mixture the reaction below was found to contain 0.171 atm of I2, 0.166 atm of Cl2 an
maw [93]

Answer: 3390

Explanation:

Since this problem already gives is the equilibrium values, all we have to do is to plug them into the formula for K_{p}.

K_{p} =\frac{[ICl]^2}{[I_{2}][Cl_{2}]  }

K_{p} =\frac{(9.81)^2}{(0.171)(0.166)} =3390

7 0
3 years ago
How many mL of 1.50 M hydrochloric acid will neutralize 25 mL of 2.00 sodium hydroxide?​
Lostsunrise [7]

Answer: The Answer is 18.7ml.

Explanation: Solved in the attached picture.

5 0
4 years ago
You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho
timama [110]

Explanation:

As per the given data, at a higher temperature, at 24^{o}C, the solution will occupy a larger volume than at 20^{o}C.

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At 20^{o}C, density of water=0.9982071 g/ml  

Therefore, \frac{concentration}{density} will be calculated as follows.

                 = \frac{C_{1}}{d_{1}}

                 = \frac{1.000 mol/L}{0.9982071 g/ml}

                 = 1.0017961 mol/g  

At 24^{o}C, density of water = 0.9972995 g/ml

Since, \frac{concentration}{density} = \frac{C_{2}}{d_{2}}

                           = \frac{C_{2}}{0.9972995}

Also,             \frac{C_{1}}{d_{1}} = \frac{C_{2}}{d_{2}}

so,                   1.0017961 mol/g = \frac{C_{2}}{0.9972995}

                      C_{2} = 1.0017961 \times 0.9972995

                                  = 0.9990907 mol/L

Therefore, in 500 ml, concentration of KNO_{3} present is calculated as follows.

             C_{2} = \frac{concentration}{volume}

               0.9990907 mol/L = \frac{concentration}{0.5 L}  

               concentration = 0.49954537 mol

Hence, mass (m'') = 0.49954537 mol \times 101.1032 g/mol = 50.5056 g       (as molar mass of KNO_{3} = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

                      m = m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}

where,          d_{air} = density of air = 0.0012 g/ml

                     d_{weight} = density of callibration weights = 8.0g/ml                      

                     d = density of weighed object

Hence, the true mass will be calculated as follows.

           True mass(m) = 50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}

             true mass(m) = 50.5268 g

                                  = 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of KNO_{3} needs to be measured.

8 0
3 years ago
4Fe(s) + 3O2(g) ⟶ 2Fe2O3(g)
oee [108]

Answer:

Iron. Fe

Explanation:

8 0
3 years ago
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what is the advantage of having folded membrane surronding the cell or within the cytoplasm or within organelles
Sunny_sXe [5.5K]
More surface area to volume ratio
6 0
3 years ago
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