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ololo11 [35]
2 years ago
5

How many grams of Hydrogen (H) would need to react with 143 grams of Oxygen (O) to make 235 grams of Water (H20)? 235

Chemistry
1 answer:
drek231 [11]2 years ago
7 0

Answer:

45 grams

Explanation:

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Water comes from different a.levels b.phases c.sources d.stages​
bezimeni [28]

Answer:

C. sources

Explanation:

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2 years ago
If an object is not moving its kinetic energy is zero.<br><br> True<br> False
Svetach [21]

Answer: This is true because kinetic energy depends on speed. If there's no speed, then there is no kinetic energy.

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3 years ago
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What happens to a brick when it is heated​
Kamila [148]

Explanation:

When a brick is heated to a temperature between 20°C and 150°C, it loses most of the water added to the clay during the preparation phase. ... When the temperature starts to rise over 600°C, chemical changes begin to occur in the clay which give the brick colour, hardness and durability.

8 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
Aluminum has a density of 2.70g/cm what is the volume occupied by a block of aluminum which weights 4.32kg
romanna [79]

Answer:

Explanation:

0.000625 cm3  

  0.625 cm3  

  1.60 cm3

4 0
3 years ago
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