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sdas [7]
3 years ago
6

During a reaction, a halogen chemically combines with a Group 1 element to form a compound. how does this reaction occur, based

on the atomic structure of each element?
A) a halogen atom loses an electron to an atom of Group 1 element
B) a halogen atom gains an election from an atom of a Group 1 element
C) a halogen atom loses two electrons to an atom of a group 1 element
D) a halogen atom shares an electron with an atom of a Group 1 element
Chemistry
2 answers:
vichka [17]3 years ago
8 0
The answer for this question would be B. Both elements, one being a Group 1 and the other being a Group 17, want to attain the octet of electrons in the outermost shell for chemical stability. The Group 1 element has one too many electrons, and the 17 is lacking one. To increase the stability of the system, the Group 1 can donate the extra electron to the Group 17 so that both can emulate a stable octet. In doing so, two ions are produced (elements that have a net charge that isn’t 0), and thus we have an “ionic bond.”

Hope this helps!
vampirchik [111]3 years ago
3 0

The right answer is B

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When 2.69 g 2.69 g of a nonelectrolyte solute is dissolved in water to make 345 mL 345 mL of solution at 26 °C, 26 °C, the solut
Gre4nikov [31]

Answer:

The molar concentration of this solution is 0.0463 mol/L

Explanation:

Step 1 : Data given

Mass of a nonelectrolyte solute = 2.69 grams

Volume of water = 345 mL = 0.345 L

Temperature = 26.0°CC = 273 + 26 = 299 K

The osmotic pressure = 863 torr

⇒ 863torr /760 = 1.13553 atm

Step 2: Calculate the molar concentration of this solution

Π = i*M*R*T

⇒with Π = the osmotic pressure = 1.13553 atm

⇒with i = the van't Hoff factor of the nonelectrolyte solute = 1

⇒with M = the molar concentration = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 299 K

1.13553 atm = 1 * M * 0.08206 L*atm/mol*K * 299 K

M = 1.13553 / (0.08206*299)

M = 0.0463 mol/L

The molar concentration of this solution is 0.0463 mol/L

5 0
2 years ago
The half life for the first order conversion of A to B is 56.6 hours. How long does it take for the concentration of A to decrea
sveta [45]

Answer:

It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.

Explanation:

A → B

Initial concentration of the reactant = x

Final concentration of reactant = 10% of x = 0.1 x

Time taken by the sample, t = ?

Formula used :

A=A_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

where,

A_o = initial concentration of reactant

A = concentration of reactant left after the time, (t)

t_{\frac{1}{2}} = half life of the first order conversion  = 56.6 hour

\lambda = rate constant

A=A_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}

Now put all the given values in this formula, we get

0.1x=x\times e^{-(\frac{0.693}{56.6 hour})\times t}

t = 188.06 hour

It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.

4 0
3 years ago
What is the volume of 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?
uysha [10]
Given:
<span> 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
volume of CL2
Solution:
Use the ideal gas law
PV = nRT
V = nRT/P
V = (2.1 moles Cl2) (0.08203 L - atm / mol - K) (273K) / (1 atm)
V = 47 L</span>
5 0
3 years ago
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

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

6 0
3 years ago
If two substances travel up the chromatography paper this implies that
Colt1911 [192]

Answer:

They are seperating

Explanation:

6 0
2 years ago
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