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olasank [31]
3 years ago
14

Boron is to the immediate left of carbon in the same row in the periodic table. Which of the following statements compares the r

adius and pull exerted by the protons of boron and carbon atoms?
A. Boron has a smaller radius and the protons in carbon exert a greater pull.
B. Boron has a larger radius and the protons in carbon exert a greater pull.
C. Boron has a smaller radius and the protons in carbon exert less pull.
D. Boron has a larger radius and the protons in carbon exert less pull.
Chemistry
2 answers:
natita [175]3 years ago
7 0

Answer:

Boron has a larger radius and the protons in carbon exert more pull.

Explanation:

Remember than elements have greater radius as they are closer to the bottom left corner, so boron would have the larger radius here. Carbon has a smaller radius, which makes it easier for the protons in carbon to exert more pull.

SOVA2 [1]3 years ago
5 0

Answer:Boron has a larger radius and the protons in carbon exert more pull.

Explanation:

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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
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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:

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Being the molar mass of the compounds:

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

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
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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

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

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