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3 years ago

2. How are these nutrients used by your body?

1 answer:

7 0

The nutrients that the body breaks down into basic units are carbohydrates, fats, and proteins. From carbohydrates comes glucose, your body's -- especially the brain's -- primary form of fuel; from fats we get glycerol and fatty acids, many of which are essential ingredients in hormones and the protective sheath in our brain that covers communicating neurons; and from proteins we get amino acids, which are the building blocks to lots of structures, including our blood, muscle, skin, organs, antibodies, hair, and fingernails.

Each of these nutrients travels down a different pathway, but all can eventually fuel the body's production of ATP (adenosine triphosphate), which is essentially our bodies' ultimate energy currency.

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What are the 10 element

maw [93]

hydrogen
oxygen
carbon
nitrogen
fluorine
silicone
boron
argon
cobalt
aluminum

hope it hepls✨

4 0

3 years ago

Is the following equation balanced? SO3 + 2H2O H2SO4 no yes

postnew [5]

Answer: No

Explanation: It's not balanced because four oxygen atoms in H2SO4, whereas there are 5 oxygen atoms in the reactants side. Also, there's more hydrogen atoms on the reactants side.

I hope this helps!

7 0

3 years ago

2. When the vapor pressure of water is 80 kpa the temperature of the

timofeeve [1]

Answer:

103

Explanation:

3 0

3 years ago

Which of the following is a physical property of water?

Kamila [148]

Water is NOT flammable.

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