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tensa zangetsu [6.8K]
3 years ago
11

Select the best answer.

Chemistry
1 answer:
Butoxors [25]3 years ago
7 0

Answer: iron

Explanation: Metals' atoms are closely bonded in a crystal lattice. In metallic bonds, the electron cloud boundaries of atoms touch each other. Many nonmetals occur as diatomic molecules. The electron clouds of these two bonded nonmetal atoms overlap .

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Each enzyme's active site is suitable for one specific type of substrate – just like a lock that has the right shape for only one specific key. Changing the shape of the active site of an enzyme will cause its reaction to slow down until the shape has changed so much that the substrate no longer fits.

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kkurt [141]

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In a long channel MOSFET, the width of the pinch-off region is assumed small relative to the length of the channel. Thus, neither the length nor the voltage across the inversion layer change beyond the pinch-off, resulting in a drain current independent of drain bias. Consequently, the drain current saturates.

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5 0
3 years ago
When 5.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of AlCl3 can be formed? (2 points) 2Al
barxatty [35]

Well, we need to find the ratio of Al to the other reactant.

Al:HCl = 1:3

--> this means that for every 1 Al used, you have to use 3 HCl.


5*3 = 15 moles of HCl needed to fully react with 5 moles of Al. Since 13<15, HCL is the limiting reactant.


The ratio of HCl:AlCl = 3:1


13/3 = 4.3333...


The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced so choice B.


Hope this helps!!! :)


5 0
4 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

3 0
3 years ago
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