Use the model of a molecule to answer the question, how many atoms of carbon does this molecule have?

Chemistry

2 answers:

snow_lady [41]3 years ago

7 0

Answer:

I think it's 4 carbon atoms

Alenkinab [10]3 years ago

6 0

Answer:

More generally, when a carbon atom is bonded to four other atoms, the molecule (or part of a molecule) will take on a tetrahedral shape similar to that of methane. (if this doesn't help sorry :<)

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For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car

Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

Na₂CO₃: 1 mole
Ca(NO₃)₂: 1 mole
CaCO₃: 1 mole
NaNO₃: 2 mole

Being the molar mass of the compounds:

Na₂CO₃: 106 g/mole
Ca(NO₃)₂: 164 g/mole
CaCO₃: 100 g/mole
NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

Na₂CO₃: 1 mole* 106 g/mole= 106 g
Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
CaCO₃: 1 mole* 100 g/mole= 100 g
NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?

$mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}$

mass of CaCO₃= 74.81 grams

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

6 0

3 years ago

Write net ionic equations for the formation of each of the precipitates observed. Express your answers as chemical equations sep

8090 [49]

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3 years ago

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Nesterboy [21]

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3 years ago

Use the following half-reactions to construct a voltaic cell:

velikii [3]

Answer: The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

Explanation:

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.