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2 years ago

Select the correct image. Which body part is vestigial in humans?

Chemistry

2 answers:

Dominik [7]2 years ago

8 0

Answer:

should be appendix, I can't see any image tho.

NNADVOKAT [17]2 years ago

6 0

Answer: Appendix

Explanation: I dont see an image though

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If you combine 27.1 g of a solute that has a molar mass of 27.1 g/mol with 100.0 g of a solvent, what is the molality of the res

baherus [9]

<u>Answer:</u> The molality of the solution is 0.1 m.

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute = 27.1 g

$M_{solute}$ = Molar mass of solute = 27.1 g/mol

$W_{solvent}$ = Mass of solvent = 100 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{27.1\times 1000}{27.1\times 100}\\\\\text{Molality of solution}=0.1m$

Hence, the molality of the solution is 0.1 m.

8 0

3 years ago

Read 2 more answers

Which of the following is not considered a behavior?

noname [10]

Mental illnesses or emotional illness or silly things like eating or dancing

3 0

3 years ago

Read 2 more answers

Consider the structures of vinegar and triglyceride and draw in any bond dipoles that exist in the molecules above using a dipol

bagirrra123 [75]

Answer:

Given molecules are vinegar and triglycerides.

Explanation:

The dipole is a vector quantity and it is heading from less electronegative atom to more electronegative atom in a polar covalent bond.

The structures and the bond dipoles in the given molecules are shown below:

5 0

3 years ago

sweet-ann [11.9K]

Answer:lmk when u have the answer

Explanation:

6 0

3 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