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Yuki888 [10]
3 years ago
9

Which figures are not significant? *

Chemistry
1 answer:
Dvinal [7]3 years ago
8 0

Answer:

leading zeroes to the left of a decimal point. and trailing zeros to the left of an invisible decimal point.

Explanation:

  • All nonzero digits are significant figures. Example: 713 has three significant figures. 18,991 has five significant figures.
  • Zeros located between two nonzero digits are significant figures. Example: 505 has three significant figures. 17,009 has five significant figures.
  • Zeros at the ends of numbers are only significant if there is a decimal point in the number. Example: 502.00 has five significant figures (5,0,2,0,0). 5020 has three significant figures (5,0,2) 100,000 has one significant figure (1)
  • Zeros that come before the first nonzero digit are not significant. Example: 0.0007 has one significant figure (7). 0.0970 has three significant figures (9,7,0)
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Nuclear fission because an atom is splitting....
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For the following reaction if you have 13.2g of CO and 42.7g of Fe2O3, which is the limiting reagent with regards to the Fe prod
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Answer: CO is a limiting reagent with regards to the Fe production.

Explanation:

Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)

Moles of CO = \frac{\text{mass of CO}}{\text{molar mass of CO}}=\frac{13.2 g}{28 g/mol}=0.4714 mol

moles of Fe_2O_3=\frac{\text{mass of} Fe_2O_3}{\text{molar mass of}Fe_2O_3}=\frac{42.7 g g}{159.7 g/mol}=0.2673 mol

According to reaction , 3 mole of CO reacts with 1 mole of Fe_2O_3 then , 0.4714 moles of CO will react with : \frac{1}{3}\times 0.4714 moles of Fe_2O_3 that is 0.1571 moles.

0.4714 moles of CO will react with 0.1571 moles of Fe_2O_3 which means that CO is present in limited amount acting as limiting reagent.

Mole remaining of Fe_2O_3 = 0.2673 mol - 0.1571 mol = 0.1102 mol

Hence, CO is a limiting reagent and  Fe_2O_3 is an excessive reagent.

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4 years ago
If an atom has an atomic mass of 24 and its atomic number is 11, how many protons will
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Answer:

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The formula relating ΔH°_rxn and enthalpies of formation (ΔH°_f) is

ΔH°_rxn = ΣΔH°_f(products) – ΣΔH°_f(reactants)

ΣΔH°_f(products) = -6(241.8) = -1450.8 kJ

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ΔH°_rxn =  (-1450.8 + 183.6) kJ = -1267.2 kJ

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