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JulsSmile [24]
3 years ago
5

Which of the following measurements is expressed to three significant figures? 7.30 × 10–7 km 0.070 mm 0.007 m 7077 mg

Chemistry
2 answers:
Aneli [31]3 years ago
5 0

Answer : The correct option is, 7.30\times 10^{-7}Km

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures are :

  • Digits from 1 to 9 are always significant and have infinite number of significant figures.
  • All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.
  • All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.
  • All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.
  • All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.
  • All zeroes used solely for spacing the decimal point are not significant. For example : 8000 has one significant figure.

As per question,

7.30\times 10^{-7}Km has 3 significant figures.

0.007 has 1 significant figure.

0.070 has 2 significant figures.

7077 has infinite significant figures.

Hence, the correct option is, 7.30\times 10^{-7}Km

trapecia [35]3 years ago
4 0
<span>7.30 x 10-7


Good luck.</span>
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Answer : The limiting reactant is Cl_2 and the theoretical yield will be 86.45 grams.

Explanation : Given,

Moles of Ti = 0.483 mole

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First we have to calculate the limiting and excess reagent.

The given balanced chemical reaction is,

Ti+2Cl_2\rightarrow TiCl_4

From the balanced reaction we conclude that

As, 2 moles of Cl_2 react with 1 mole of Ti

So, 0.911 moles of Cl_2 react with \frac{0.911}{2}=0.455 moles of Ti

From this we conclude that, Ti is an excess reagent because the given moles are greater than the required moles and Cl_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of TiCl_4.

As, 2 moles of Cl_2 react to give 1 moles of TiCl_4

So, 0.911 moles of Cl_2 react to give \frac{0.911}{2}=0.455 moles of TiCl_4

Now we have to calculate the mass of TiCl_4.

\text{Mass of }TiCl_4=\text{Moles of }TiCl_4\times \text{Molar mass of }TiCl_4

\text{Mass of }TiCl_4=(0.455mole)\times (190g/mole)=86.45g

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