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-Dominant- [34]
3 years ago
12

How many liters of carbon dioxide will be produced at STP if 3.56 g calcium carbonate reacts completely with carbon dioxide? CaC

O3 --> CaO + CO2
Chemistry
1 answer:
sp2606 [1]3 years ago
4 0

Answer:

V = 0.798 L

Explanation:

Hello there!

In this case, for this gas stoichiometry problem, we first need to compute the moles of carbon dioxide via stoichiometry and the molar mass of starting calcium carbonate:

3.56gCaCO_3*\frac{1molCaCO_3}{100gCaCO_3} *\frac{1molCO_2}{1molCaCO_3} =0.0356molCO_2

Next, we use the ideal gas equation for computing the volume, by bearing to mind that the STP conditions stand for a pressure of 1 atm and a temperature of 273.15 K:

PV=nRT\\\\V=\frac{nRT}{P}\\\\V=\frac{0.0356mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm} \\\\V=0.798L

Best regards!

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A cube of an unknown metal measures 0.200 cm on one side. The mass of the cube is 52 mg. Which of the following is most likely t
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Answer:

Option E. Zirconium

Explanation:

From the question given above, the following data were obtained:

Length of side (L) of cube = 0.2 cm

Mass (m) of cube = 52 mg

Name of the unknown metal =?

Next, we shall determine the volume of the cube. This can be obtained as follow:

Length of side (L) of cube = 0.2 cm

Volume (V) of the cube =?

V = L³

V = 0.2³

V = 0.008 cm³

Next, we shall convert 52 mg to g. This can be obtained as follow:

1000 mg = 1 g

Therefore,

52 mg = 52 mg × 1 g / 1000 mg

52 mg = 0.052 g

Thus, 52 mg is equivalent to 0.052 g.

Next, we shall determine the density of the unknown metal. This can be obtained as follow:

Mass = 0.052 g.

Volume = 0.008 cm³

Density =?

Density = mass / volume

Density = 0.052 / 0.008

Density of the unknown metal = 6.5 g/cm³

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7 0
3 years ago
2. In the reaction NO + NO2 ⇌ N2O3, an experiment finds equilibrium concentrations of [NO] = 3.8 M, [NO2] = 3.9 M, and [N2O3] =
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Kc= concentration  of  product  divided  by  concentration   of  reactant
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Kc =(N2O3)  /  (No)(NO2)

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4 0
3 years ago
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What is he empirical formula for the compound that is of 1.85 moles of nitrogen and 4.63 miles of oxygen
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The empirical formula is N₂O₅.

The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.  

The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of N:O</em>.  

I like to summarize the calculations in a table.  

<u>Element</u> <u>Moles</u>  <u>Ratio¹ </u>  <u> ×2²  </u>  <u>Integers</u>³

     N        1.85    1             2             2

     O        4.63    2.503   5.005     5

¹To get the molar ratio, you divide each number of moles by the smallest number (1.85).

²Multiply these values by a number (2) that makes the numbers in the ratio close to integers.

³Round off the number in the ratio to integers (2 and 5).

The empirical formula is N₂O₅.

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3 years ago
The expected o-c-h angle in this molecule is degrees. the expected hybridization at the central carbon is
WINSTONCH [101]

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Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

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Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible

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Explanation:

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