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qwelly [4]
3 years ago
10

0.0136 g + 2.70 × 10-4 g - 4.21 × 10-3 g = ? How many digits to the right of the decimal point should be used to report the resu

lt?
Chemistry
2 answers:
cestrela7 [59]3 years ago
5 0

Answer:

You should use 5 digits to the right of the decimal point.

Explanation:

To be precise in the mathematical operation you have to conserve all significant digits .

Replace the scientific notation for its equivalent in decimal notation:

2.70 x 10-4 g = 0.00027  g

4.21 x 10-3  g = 0.00421  g

Then, compute the addition and subtraction :

0.01360  g

+  

0.00027  g

-  

0.00421  g

-------------

0.00966 g = 9.66 x 10-3 g

That means you should use 5 digits to the right of the decimal point, which are the biggest amount of digits necessary to represent  2 of the numbers involved in the mathematical operation.

ki77a [65]3 years ago
4 0

Consider the following equation:0.0136 g + 2.70 × 10-4 g - 4.21 × 10-3 g  = ? How many digits to the right of the decimal point should be used to report the result? =2, Choose the correct answer=9.66 × 10^-3 g, Which metric unit would be the best choice to report the result?=mg

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Explanation:

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3 years ago
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posledela

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of HNO_3 = \frac{15.5g}{63g/mol\times 9.5L}=0.026M

Equilibrium concentration of NO = \frac{16.6g}{30g/mol\times 9.5L}=0.058M

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Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c  

For the given chemical reaction:

2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)

The expression for K_c is written as:

K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}

K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}

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Thus  the value of the equilibrium constant Kc for this reaction is 3.72

5 0
3 years ago
Where does the equilibrium point occur in a reaction system?
Ymorist [56]
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