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lina2011 [118]
3 years ago
6

Which shows zero as significant digits?check all that apply.

Chemistry
1 answer:
DiKsa [7]3 years ago
8 0
- any zeroes between 2 non-zeroes
- any zeroes after non-zeroes in a number without a decimal point are sig figs



These are NOT:

- any zeroes before non-zeroes
- any zeroes after non-zeroes in a number without a decimal point
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When iron reacts with oxygen, it forms iron oxide, or rust.
Feliz [49]

Answer:

Mass of iron oxide: 79.85 g  

Explanation:

Given data:

Mass of iron = 112 g

Mass of O₂ = 24 g

Mass of iron oxide formed = ?

Solution:

4Fe + 3O₂         →       2Fe₂O₃

Number of moles of Fe:

Number of moles = mass/ molar mass

Number of moles = 112 g/ 55.85 g/mol

Number of moles =  2 mol

Number of moles of O₂:

Number of moles = mass/ molar mass

Number of moles = 24 g/ 32 g/mol

Number of moles =  0.75 mol

Now we will compare the moles of oxygen and iron with iron oxide.

                 Fe           :             Fe₂O₃

                 4             :                2

                 2             :                2/4×2 =  1

                O₂            :             Fe₂O₃

                 3             :                2

                 0.75       :                2/3×0.75 =  0.5

The number of moles of iron oxide formed by oxygen are less thus it will be limiting reactant.

Mass of iron oxide:

Mass = number of moles × molar mass

Mass = 0.5 mol  × 159.69 g/mol

Mass = 79.85 g  

3 0
3 years ago
2.
Dmitry_Shevchenko [17]

Answer:

Atomic number

Explanation:

6 0
3 years ago
Consider the reactionI2(g) + Cl2(g)2ICl(g)Using standard thermodynamic data at 298K, calculate the entropy change for the surrou
Georgia [21]

We know,

\Delta H_{I_2(g)}=62.438\ KJ/mol\\\\\Delta H_{Cl_2(g)}= 0.0\ KJ/mol\\\\\Delta H_{ICl(g)}=17.78\ KJ/mol

For given reaction, I_2(g)+Cl_2(g)\ -->\ 2ICl(g)

\Delta H_{rxn}=2\Delta H_{ICl(g)}-\Delta H_{I_2(g)}-\Delta H_{Cl_2(g)}\\\\\Delta H_{rxn}=2(17.78)-0-62.438\ KJ/mol\\\\\Delta H_{rxn}=-26.878\ KJ/mol

For , 2.41 moles of I_2 :

\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ

We know :

\Delta S = -\dfrac{\Delta H_{rxn}}{T}\\\\\Delta S = -\dfrac{-64.78}{298}\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K

Hence, this is the required solution.

7 0
3 years ago
What does “like dissolves like” mean?
Tems11 [23]

Answer:

c  solvents dissolve chemicals with the same polarity ( ex. both are polar)

Explanation:

Like dissolves like is one of the central rule that guides the solubility of one substance in another.

  • It fully suggests substance having the same nature as in polarity-wise will dissolve one another.
  • For example, water is a polar liquid, it will dissolve table salt because it i also polar.
  • Water cannot dissolve oil because oil is non-polar.
4 0
3 years ago
The half-life for the radioactive decay of C-14C-14 is 5730 years. You may want to reference (Pages 598 - 605) Section 14.5 whil
KIM [24]

<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life of the reaction = 5730 years

Putting values in above equation, we get:

k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant  = 1.21\times 10^{-4}yr^{-1}

t = time taken for decay process = ? yr

[A_o] = initial amount of the sample = 100 grams

[A] = amount left after decay process =  (100 - 25) = 75 grams

Putting values in above equation, we get:

1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

8 0
3 years ago
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