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ASHA 777 [7]
2 years ago
15

Calculate the density of a substance with a mass of 0.0500g and volume of 6.40mL, round to the correct number of significant num

bers.
Chemistry
1 answer:
lozanna [386]2 years ago
5 0
Density=mass/volume
Mass = 0.0500g
Volume = 6.40mL

0.0500g/6.40mL = 0.0078g/mL
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PLS HELP QUICK ALOTTT OF POINTS
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Answer:

\boxed {\boxed {\sf 0.80 \ mol\ F}}

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

Flip the ratio so the units of atoms of fluorine cancel each other out.

4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}

4.8 \times 10^{23}  *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }

Condense into 1 fraction.

\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F

Divide.

0.7970773829 \ mol \ F

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

0.80 \ mol \ F

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>

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True

Explanation:

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Yeast and other organisms can convert glucose (C6H12O6) to ethanol (CH3CH2OH) by a process called alchoholic fermentation. The n
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Answer:

8.37 grams

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C₆H₁₂O₆     ⇒   2 C₂H₅OH (l) + 2 CO₂ (g)

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From the ideal gas law we can determine the number of moles that the 2.25 L represent.

From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.

PV = nRT    ⇒ n = PV/RT

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Moles glucose required:

0.093 mol CO₂  x  ( 1 mol C₆H₁₂O₆   / 2 mol CO₂ ) =  0.046 mol C₆H₁₂O₆

The molar mass of glucose is 180.16 g/mol, then the mass required is

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