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Ganezh [65]
3 years ago
15

Propan-2-ol added to butanoic acid

Chemistry
1 answer:
nasty-shy [4]3 years ago
6 0
Dhidndjd dheudjdn dhdidjdndb
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Aleksandr-060686 [28]
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moles * molar mass = mass
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PLEASE HELP ILL GIVE U BRAINLIEST<br> What's something you learned about the periodic table
valina [46]

Answer:

The periodic table of elements puts all the known elements into groups with similar properties. This makes it an important tool for chemists, nanotechnologists, and other scientists. If you get to understand the periodic table and learn to use it, you'll be able to predict how chemicals will behave.

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5 0
3 years ago
What is the energy required to change a spherical drop of water to five smaller spherical drops of equal size? At room temperatu
pav-90 [236]
T<span>his is a straightforward question related to the surface energy of the droplet. </span>

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
3 0
3 years ago
Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits<br> and put the units.
marta [7]

Answer:

Approximately 53.3\; \rm g.

Explanation:

Lookup Avogadro's Number: N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1} (three significant figures.)

Lookup the relative atomic mass of \rm H, \rm S, and \rm O on a modern periodic table:

  • \rm H: 1.008.
  • \rm S: 32.06.
  • \rm O: 15.999.

(For example, the relative atomic mass of \rm H is 1.008 means that the mass of one mole of \rm H\! atoms would be approximately 1.008\! grams on average.)

The question counted the number of \rm H_2SO_4 molecules without using any unit. Avogadro's Number N_{\rm A} helps convert the unit of that count to moles.

Each mole of \rm H_2SO_4 molecules includes exactly (1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23} of these \rm H_2SO_4 \! molecules.

3.27 \times 10^{23} \rm H_2SO_4 molecules would correspond to \displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of \rm H_2SO_4 gives the mass of each mole of \rm H_2SO_4\! molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the mass of approximately 0.541389\; \rm mol of \rm H_2SO_4:

\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}.

(Rounded to three significant figures.)

6 0
3 years ago
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