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kati45 [8]
2 years ago
9

(23.4m)(14m) rounded off to the correct number of significant figures

Chemistry
1 answer:
VARVARA [1.3K]2 years ago
5 0

The value of the given expression up to correct significant figures is 330 m^2.

  • The rule that applies for the multiplication and division is :

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

  • The rule that applies for the addition and subtraction is :

The least precise number present after the decimal point determines the number of significant figures in the answer.

Given:

The expression: (23.4m)(14m)

To find:

The answer up to correct a number of significant figures.

Solution

Number of significant figures in '23.4' = 3

Number of significant figures in '14' = 22

=(23.4m)(14m)\\=(23.4m)\times (14m)\\=327.6 m^2

The number with the least significant figures is '14' is 2, so the answer will be reported up to 2 significant figures:

=327.6 m^2\approx 330 m^2

Number of significant figures in '330' = 3

The value of the given expression up to correct significant figures is 330 m^2.

Learn more about significant figures here:

brainly.com/question/11904364?referrer=searchResults

brainly.com/question/17177568?referrer=searchResults

 

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Answer:

All are correct

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l=0 (sphere - s-orbital)

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2) The magnetic quatum number, ml relates to the number of orbitals within a subshell then it is related with l, taking values form -l to l incluing 0.

For l=0 (s-orbital) ml=0

For l=1 (p-orbital) ml=1,0,-1

For l=2 (d-orbital) ml=2,1,0,-1,-2

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2 years ago
Value of Δ H ∘ rxn for the equation NH 3 ( g ) + 2 O 2 ( g ) ⟶ HNO3 ( g ) + H 2 O ( g )
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Answer: - 894.6 kJ/mol.

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Hess law is states that the changes in enthalpies in a chemical reactions is independent of the pathway between the initial and final states.

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We are to find the Value of ΔH°(rxn) for the equation:

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ----------------------------------(**).

From the series of equations given;

==> 4NH3 (g) + 5O2 (g) -------->4 NO(g) + 6H2O (l). ∆H = -1166.0 kJ/mol.--------------------------------------(1).

===> 2NO(g) + O2 (g) ------> 2NO2 (g). ∆H = -116.2 kJ/mol.---------------(2).

===> 3NO2 (g) + H2O (l) ---------> 2HNO3 (aq) + NO (g). ∆H = -137.3 kJ/mol.-------------------------------------(3).

The first thing to do is to multiply equation (2) by 3. Also, multiply equation (3) by 2. This will give us equation (4) and (5) respectively.

6NO + 3 O2 ----------------> 6NO2. ∆H= 3 × (-116.2 kJ/mol) = -348.6 kJ/mol. ------------------------------------(4).

6NO2 + 2 H2O ----------------> 4HNO3 + 2 NO. ∆H= 2 × (-137.3kj/mol) = -274.6 kJ/mol ---------------------------(5).

Next, add equations (4) and (5) to give;

4NO +3O2 +2H2O -------------> 4HNO3. ∆H = -623.2 kJ/mol. -----(6).

Add this equation to the equation (1) from above, we have;

4NH3 + 8O2 --------------> 4HNO3 + 4H2O. ∆H= -1789.2 kJ/mol. --------(7).

Then, divide the equation (7) above by 2 to give us back the equation (**).

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ∆H= -894.6 kJ/mol.

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H  O  W  E  V  E  R

the constant formula you can use for solving and simplifying equations such as these, you can simply look up the metric system online and figure one out. hoped this helped!
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