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nikdorinn [45]
3 years ago
9

6. when multiplying and dividing measured quantities, the number of significant figures in the result should be equal to the num

ber significant figures in___. a) all of the measurements b) at least and most precise measurements c) the most precise measurement d) the least precise measurement
Chemistry
2 answers:
8090 [49]3 years ago
4 0

The correct option is D.

Significant figure refers to the number of digits that are used to report a measured quantity. The amount of a significant number in a measurement always equal to the number of digits that are known with some degree of confidence and the last number, which is usually an estimate. When carrying out calculations with significant figures measurements, one must be careful to report the number of significant figure to reflect the least precise measurement.

Vlad [161]3 years ago
4 0
The least precise measurement.

When figuring out how many significant figures to go to, you always use the least amount of sig figs in the question so the least sig figs means the least precise.
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k0ka [10]

Answer:

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Explanation:

We know that sharing of electrons form covalent bonds.

If we look upon K,L,M ,N shells of the carbon and hydrogen atoms.

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And Carbon on the other hand is having 2 electrons in K shell and 4 electrons in L shell.

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And Hydrogen requires  1 electron to complete its doublet.

Alkane general formula C_{n}H_{2n+2}

For ethane n=2

C_{2}H_{4+2} = C_{2}H_{6}

1 Carbon atom is shared by  3 Hydrogen.

The remaining one electron (4-3)=1 of carbon will be shared with another carbon atom.

An image of the sharing of electrons attached below,

Hence we have only  1 covalent bond between the two.

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3 years ago
Suppose 215 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?
charle [14.2K]

Answer:

The answer is "41.23 \ L\  N_2"

Explanation:

2 NO_3^{-} + 10 e^{-} + 12 H^{+} \longrightarrow N_2 + 6 H_2O\\\\= \frac{( 215 \ g \ NO_3^{-})}{(62.0049  \frac{\ g NO_3^{-}}{mol})} \times  \frac{(1 \ mol \ N_2}{ 2 \ mol \ NO_3^{-})}\\\\

=3.46746789 \times 0.5\\\\= 1.733 \ mol \ N_2 \\\\\to V = \frac{nRT}{P} \\\\= (1.733 \ mol) \times (0.08205746 \frac{L\ atm}{Kmol}) \times \frac{ (17 + 273) K}{(1.00 atm)}\\\\= 41.23

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2 years ago
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