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3 years ago

you are given a sample containing 0.70 moles of a substance. How many atoms are in the sample if the substance is zinc?

1 answer:

7 0

1 mole ------------- 6.02 x 10²³ atoms
0.70 moles -------- ( atoms zinc )

atoms platinum = 0.70 x ( 6.02 x 10²³ ) / 1

 = 4.2 x 10²³ atoms zinc

hope this helps!

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WORTH 10 POINTS!! Strawberry plants produce offspring through asexual reproduction. Which statement is true about strawberry pla

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Answer:

They are the same as one parent :)

Explanation: that is why we have more yellow corn than black corn because people planted only yellow corn to make more yellow corn :D

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2 years ago

You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution

vfiekz [6]

Answer:

450. g of 0.173 % KCN solution contains 779 mg of KCN.

Explanation:

Mass of the solution = m

Mass of the KCN in solution = 779 mg

Mass by mass percentage of KCN solution = 0.173%

$(m/m)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

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3 years ago

How many hydrogen atoms are in a cycloalkane with 8 carbon atoms?

frosja888 [35]

Cycloalkanes are those saturated organic compounds which exist in the form of Rings. Their Hydrogen Deficiency Index in one. The General formula for cycloalkanes is,
CnH2n
When number of Carbons = 8
Then
C₈H₂₍₈₎

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Result:
Cycloalkane containing 8 carbon atoms has 16 hydrogen atoms.

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3 years ago

Atoms with greatly different electronegativity values are expected to form _____. no bonds ionic bonds none of these covalent bo

Ad libitum [116K]

Atoms with greatly different electronegativity values are expected to form ionic bonds

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3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

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3 years ago