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Sever21 [200]
2 years ago
4

Which of the following groups contains reactive elements that need to gain only one electron to have a full outer energy level?

Chemistry
1 answer:
kumpel [21]2 years ago
5 0
Alkali metals (group 1A elements)
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g A 0.25 mol sample of HBr is added to a 1.00 L buffer solution consisting of 0.68 M HCN and 0.35 M NaCN. Identify all species t
salantis [7]

Answer:

See explaination for details and diagram

Explanation:

We can say that Chemical reaction is a process in which one or more substances, the reactants, are converted to one or more different substances, the products. Substances are either chemical elements or compounds.

A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products.

See attachment for the buffer of the chemical reaction.

7 0
2 years ago
Calculate the freezing point of a 0.100 m aqueous solution of k2so4, taking interionic attractions into consideration by using t
Leokris [45]
Ionic  salt  dissociate  completely  in  water   particularly in  water at  low  concentration.
 The  molal  freezing point  of  depression constant  for  water is 1.85kg/k/mol
therefore depression  of  freezing  point =1.853  x 0.100  x  2.32=0.429  degrees  celsius
hence  solute  freeze at - 0.429  degree  celsius
3 0
2 years ago
A line in the Brackett series of hydrogen has a wavelength of 1945 nm. From what state did the electron originate?
fiasKO [112]
Rydberg Eqn is given as:
 1/λ = R [1/n1^2 - 1/n2^2] 
<span>Where λ is the wavelength of the light; 2626 nm = 2.626×10^-6 m </span>
<span>R is the Rydberg constant: R = 1.09737×10^7 m-1 </span>
<span>From Brackett series n1 = 4 </span>
<span>Hence 1/(2.626×10^-6 ) = 1.09737× 10^7 [1/4^2 – 1/n2^2] </span>
<span>Some rearranging and collecting up terms: </span>
<span>1 = (2.626×10^-6)×(1.09737× 10^7)[1/16 -1/n2^2] </span>
<span>1= 28.82[1/16 – 1/n2^2] </span>
<span>28.82/n^2 = 1.8011 – 1 = 0.8011 </span>
<span>n^2 = 28.82/0.8011 = 35.98 </span>
<span>n = √(35.98) = 6</span>
4 0
2 years ago
Read 2 more answers
Fluorine gas is placed in contact with calcium metal at high temperatures to produce calcium fluoride powder. What is the formul
Mars2501 [29]

Answer:

Upper F subscript 2 (g) plus upper C a (s) right arrow with delta above upper C a upper F subscript 2 (s).

Explanation:

This is a chemical reaction problem.

In expressing any chemical reaction, we need to understand that there are reactants and products.

  • The reactants are the species on the left hand side that are combining.
  • The products are the species on the right hand side that are formed.
  • Every chemical reaction is obeys the law of conservation of matter i.e equal number of matter on both sides.

Using the statement of this problem, we can deduce that;

 Reactants are Fluorine gas and Calcium metal

  Product is Calcium Fluoride

Note: A metal is a solid(s) and powder is a solid(s). A gas is denoted as (g). They depict the state of the species reacting.

                    F₂_{(g)}     +    Ca_{(s)}               →           CaF₂_{(s)}

We can see that equal number of atoms are on both sides of the expression.

7 0
2 years ago
Read 2 more answers
A sample of gas contains 0.1800 mol of CO(g) and 0.1800 mol of NO(g) and occupies a volume of 23.2 L. The following reaction tak
baherus [9]

Answer:

The volume of the sample is 17.4L

Explanation:

The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:

0.1800mol + 0.1800mol reactants =

0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.

Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:

V1n2 = V2n1

<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>

Replacing:

V1 = 23.2L

n2 = 0.2700 moles

V2 = ??

n1 = 0.3600 moles

23.2L*0.2700mol = V2*0.3600moles

17.4L = V2

<h3>The volume of the sample is 17.4L</h3>
8 0
1 year ago
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