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IrinaK [193]
3 years ago
9

Identify each element below, and give the symbols of the other elements in its group:

Chemistry
1 answer:
Misha Larkins [42]3 years ago
4 0

Answer:

Answer in explanation

Explanation:

a. Boron , element 5

Helium has 2 electrons, add to the other 3 to give 5.

Other group members are : Aluminum Al, Gallium Ga, Indium In , Thallium Tl and Nihonium Nh

b. Sulphur, element 16

Neon is 10 , add other 6 electrons to make 16

Other group members are: Oxygen O, selenium Se , Tellurium Te and Polonium Po

c. Lanthanum, element 57

Xenon is 54, add the other 3 electrons to give 57.

Other elements in group : Scandium Sc , Yttrium Y , Actinium Ac, Lutetium Lu and/or Lawrencium Lr

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One part nitrogen gas combines with one part oxygen gas to form how many part(s) dinitrogen monoxide (nitric oxide)?
____ [38]

Answer : The one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

When nitrogen gas combines with oxygen gas then it react to give dinitrogen monoxide or nitrous oxide.

The balance chemical reaction will be:

2N_2(g)+O_2(g)\rightarrow 2N_2O(g)

By the stoichiometry we can say that, 2 parts of nitrogen gas combines with 1 part of oxygen gas to give 2 parts of dinitrogen monoxide or nitrous oxide.

First we have to determine the limiting reagent.

From the reaction we conclude that,

As, 2 moles of nitrogen gas combine with 1 mole of oxygen gas

So, 1 moles of nitrogen gas combine with 0.5 mole of oxygen gas

It means that, oxygen gas is an excess reagent because the given moles are greater than the required moles and nitrogen gas is a limiting reagent and it limits the formation of product.

Now we have to determine the moles of dinitrogen monoxide.

As, 2 moles of nitrogen gas combine to give 2 mole of dinitrogen monoxide

So, 1 mole of nitrogen gas combine to give 1 mole of dinitrogen monoxide

Thus, the one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

7 0
3 years ago
how much heat energy is needed to raise the temperature of 78.4g of aluminum from 19.4 degrees c to 98.6 degrees c.
Iteru [2.4K]

 The heat that is needed  to raise the   temperature  of 78.4 g of aluminium from 19.4 °c to 98.6°c  is    5600.77 j

 <u><em>calculation</em></u>

Heat(Q) = mass(M) x  specific heat capacity (C) x change in temperature(ΔT)


where;

Q=?

M = 78. 4 g

C=0.902 j/g/c

ΔT=98.6°c -19.4°c =79.2°c

Q is therefore = 78.4 g  x 0.902 j/g/c  x 79.2°c  =5600.77 j

7 0
3 years ago
The average atomic mass recorded on the periodic table for krypton is 83.80 u. This indicates that the most abundant isotope of
Trava [24]
Krypton-84 is the most abundant type of Krypton. The answer is C.

 If you need this to be explained, I  will do my best it's a bit difficult to say how to find it out, but I will if you need me to.
7 0
3 years ago
Why do scientists look for patterns in the world?
PSYCHO15rus [73]

Answer:

B. Patterns can help explain observations.

Explanation:

Hope this helps

6 0
3 years ago
Now they feel it is best to have you identify an unknown gas based on its properties. Suppose 0.508 g of a gas occupies a volume
pishuonlain [190]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

             mass = 0.508 g,               Volume = 0.175 L

             Temperature = (25 + 273) K = 298 K,       P = 1 atm

As per the ideal gas law, PV = nRT.

where,  n = no. of moles = \frac{mass}{\text{molar mass}}

Hence, putting all the given values into the ideal gas equation as follows.

               PV = \frac{mass}{\text{molar mass}} \times RT            

           1 atm \times 0.175 L = \frac{0.508 g}{\text{molar mass}} \times 0.0821 L atm/ K mol \times 298 K  

                            = 71.02 g

As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of Cl_{2} is 70.8 g/mol or 71 g/mol (approx).

Thus, we can conclude that the gas is most likely chlorine.

4 0
3 years ago
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