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3 years ago

Which of these is a carbohydrate?

Chemistry

1 answer:

Paladinen [302]3 years ago

6 0

Answer:

C6H12O6

Glucose

Explanation:

Glucose is a monocarbohydrate

carbohydrates always have only C, H, and O

brainliest plz

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Which redox reaction would most likely occur if zinc and copper metal were

sweet-ann [11.9K]

Answer:

Explanation:

b/c when copper and zinc metal are addedto solution,then the solution will be consider under redox reaction

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3 years ago

In The Lord of the Rings, Frodo and Sam use a rope to rappel down a cliff. The rope was 30 ells long. How high was the cliff in

Tcecarenko [31]

Answer:

324 meters

Explanation:

I don't really have an explanation

7 0

3 years ago

Calculate the cell potential, Ecell, of a concentration cell containing a 0.1 M solution of Zn2 and a 1.0 M solution of Zn2

Agata [3.3K]

Answer:

The cell potential is 0.0296V

Explanation:

Please see the attachment below

3 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

Use the chart to determine which type of bond is formed between potassium (K) and chlorine (Cl).

iris [78.8K]

Potassium and Chloride forms an ionic bond.
(K+) + (Cl-) = KCl

Potassium is under Group IA (Alkali Metal), wherein elements under this group can easily lose electrons.

Chlorine is under Group VII (Halogens), in which these elements can gain electrons easily.

The inner shell electrons on potassium will merge with the outer shell of electrons of chlorine to make potassium chloride.

6 0

3 years ago

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