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Mekhanik [1.2K]
3 years ago
11

PLEASE HELP!!!

Chemistry
1 answer:
neonofarm [45]3 years ago
8 0

Answer:

Cl_2+2NaI\rightarrow I_2+2NaCl

Explanation:

Hello there!

In this case, according to the described chemical reaction, Cl2 replaces iodine in NaI in order to produce I2 and NaCl:

Cl_2+NaI\rightarrow I_2+NaCl

It is possible to realize how chlorine replaces iodine in agreement with the single displacement reaction. Moreover, since chlorine and iodine atoms are not correctly balanced, we add a 2 in front of both NaI and NaCl in order to do so:

Cl_2+2NaI\rightarrow I_2+2NaCl

Best regards!

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The density of water is 1.00 g/ml at 4°c. how many water molecules are present in 2.56 ml of water at this temperature?
o-na [289]

Answer:

             8.55 × 10²² Molecules

Solution:

Step 1: Calculate the Mass of Water as;

                       Density  =  Mass ÷ Volume

Solving for Mass,

                       Mass  =  Density × Volume

Putting values,

                       Mass  =  1 g.mL⁻¹ × 2.56 mL

                       Mass  =  2.56 g

Step 2: Calculate Moles of Water as,

                       Moles  =  Mass ÷ M.Mass

Putting values,

                       Moles  =  2.56 g ÷ 18 g.mol⁻¹

                       Moles  =  0.142 mol

Step 3: Calculate Number of Molecules of Water as,

                       # of Molecules  =  Moles × 6.022 × 10²³

Putting value of mole,

                       # of Molecules  =  0.142 × 6.022 × 10²³

                       # of Molecules  =  8.55 × 10²² Molecules

5 0
3 years ago
an iron bar weighed 664 g. after the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to ru
DaniilM [7]

The final mass of the iron bar and the rust when the weight of the bar was 664g and the bar had been standing in moist air for a month, will be 688.456g.

One-eighth of the iron bar turns to rust, which means the mass of the iron bar that has undergone rusting= 664/8= 83g

Thus, the mass of pure iron left after rusting has taken place will be= (664-83)g= 581g.

The equation of rusting is as follows

4Fe + 3O2 + 6H2O → 4Fe(OH)3

Molecular weight of iron is 55.84g, thus iron will undergo the reaction= 83/55.84= 1.486 mole

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4 moles of iron produce 4 moles of rust or iron oxide

Thus, 1.486 mole of iron will produce 1.486 mole iron oxide

Mass of iron oxide= 159.68/ 1.486

Mass of rust           = 107.456g

Therefore, mass of iron bar= (581+ 107.456)g

                                            = 688.456g

To learn more about rusting, click

brainly.com/question/18376414

#SPJ4

5 0
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