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guajiro [1.7K]
3 years ago
7

The numbers used in a balanced chemical equation to represent the number of particles of

Chemistry
1 answer:
Ede4ka [16]3 years ago
7 0

Two: Coefficients are the numbers in front of each formula. If no number is shown, a one is understood. Three: The coefficients tell us how many molecules (moles) of each reactant used and how many molecules (moles) of each product made.

You might be interested in
1.6x10^23 lead atoms. Find the weight in grams
Ber [7]
Moles of lead(Pb) = 1.6x10^23/6.02x10^23 = 0.265 moles.

Weight of lead = moles x atomic weight of lead
                         =  0.265x207.2
                         =  54.908 grams.

Hope this helps!
5 0
3 years ago
rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o
SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)


Initial important note:


Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.


1) Air:


Source: internet


Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.


2) Exhaled air:


Source: internet.


Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂


3) Air from the decomposition of H₂O₂.


In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:


i) Chemical Equation:


H₂O₂ (g) → H₂ (g) + O₂ (g)


ii) mole ratio of the products 1 mol H₂ : 1 mol O₂


iii) convert moles into mass (grams)


1 mol H₂ × 2 × 1.008 g/mol = 2.016 g


1 mol O₂ × 2 × 15.999 g/mol = 31.998 g


Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%



4) Air from the decomposition of NaHCO₃:


i) chemical equation:


2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)


ii) mole ratio: take into account only the gases in the products:


1 mol CO₂ (g) : 1 mol H₂O


iii) mass in grams


CO₂: molar mass ia approximately 44.01 g/mol


H₂O: molar mass is approximately 18.02 g/mol


iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.


5) The result is:


H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

7 0
3 years ago
Read 2 more answers
Which subscripts would properly complete the formula unit Al_N_?
MakcuM [25]
I believe that it would be Al1N1.
4 0
3 years ago
Could someone help me with this?
Varvara68 [4.7K]

Solving part-1 only

#1

KMnO_4

  • Transition metal is Manganese (Mn)

#2

Actually it's the oxidation number of Mn

Let's find how?

\\ \tt\Rrightarrow x+1+4(-2)=0

\\ \tt\Rrightarrow x+1-8=0

\\ \tt\Rrightarrow x-7=0

\\ \tt\Rrightarrow x=+7

  • x is the oxidation number

#3

  • Purple as per the color of potassium permanganate

#4

\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}

7 0
2 years ago
Homolysis, or homolytic bond dissociation, produces a very specific type of product under certain reaction conditions. In Part 1
vekshin1

Answer:

1) R₃CO , H, H₃C, a carbon free radical

2) high temperature, ultraviolet irradiation

Explanation:

1) Homolysis leads to the formation of free radicals (species having a free electron). Thus, answer is :

R₃CO

H

H₃C

a carbon free radical

2) Homolysis require high temperature, ultraviolet irradiation.

5 0
3 years ago
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