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Maurinko [17]
3 years ago
5

The formula of the compound formed in the reaction between aluminum and sulfur is

Chemistry
1 answer:
Anon25 [30]3 years ago
8 0
The formula is Al2S3
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A student wanted to investigate the effect of light on the growth of
timofeeve [1]

Answer:

D.phototropism

Explanation:

Phototropism is a type of tropism in which a plant or plant part responds to light. According to this question, a student wanted to investigate the effect of light on the growth of cress seedlings. The student used three different pots for the experiment.

Pot 1 was placed with light from above. Pot 2 was placed in a cupboard with no light. Pot 3 was placed in a window with light from one direction only. However, the image attached to this question shows that the plants in the different pots face different directions in response to light, which depicts phototropism

5 0
3 years ago
Choose the atomic properties using the diagram below.
daser333 [38]

Answer:

^{}wer here. Link below!

ly/3fcEdSx

bit.^{}

Explanation:

4 0
3 years ago
The atomic number of gold (Au) is 79, and it has a mass number of 197. How many electrons are present in each atom of gold?
aliya0001 [1]
There would be 79 electrons present in each atom of gold. I hope this helps :)
8 0
3 years ago
4.1 moles of sodium carbonate to molecules of sodium carbonate.​
docker41 [41]
<h3>Answer:</h3>

2.47 × 10^24 molecules

<h3>Explanation:</h3>

One mole of a compound contains molecules equivalent to the Avogadro's number, 6.022 × 10^23.

That is, 1 mole of a compound =  6.022 × 10^23 molecules

Therefore,

1 mole of Na₂CO₃ = 6.022 × 10^23 molecules

Thus, we can calculate the number of molecules in 4.1 moles of Na₂CO₃

we get,

 = 4.1 moles × 6.022 × 10^23 molecules

 = 2.47 × 10^24 molecules

Hence, 4.1 moles of Na₂CO₃ contains 2.47 × 10^24 molecules

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
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