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Anna71 [15]
1 year ago
15

How many protons, neutrons, and electrons for b+5

Chemistry
1 answer:
Aliun [14]1 year ago
8 0

Explanation:

It has 5 protons,6 neutrons, 5 electrons.

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In a test of an automobile engine 1.00 L of octane (702 g) is burned, but only 1.84 kg of carbon dioxide is produced. What is th
Reptile [31]

Answer:

The % yield of CO2 is 85.05 %

Explanation:

Step 1: Data given

Mass of octane = 702 grams

Molar mass octane = 114.23 g/mol

Mass CO2 =1.84 kg = 1840 grams

Molar mass of CO2

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles of octane

Moles octane = mass octane / molar mass octane

Moles octane = 702.0 grams / 114.23 g/mol

Moles octane = 6.145 moles

Step 4: Calculate moles of CO2

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

For 6.145 moles octane we'll have 8*6.145 moles =49.16 moles

Step 5: Calculate mass of CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 49.16 moles * 44.01 g/mol

Mass CO2 = 2163.5 grams

Step 6: Calculate % yield of carbon dioxide

% yield = (actual yield / theoretical yield)*100%

% yield = (1840/2163.5)*100%

% yield = 85.05 %

The % yield of CO2 is 85.05 %

4 0
3 years ago
According to the following balanced equation, 2 formula units of Iron (III) Oxide (Fe2O3) can be formed by reacting 4 atoms of i
-Dominant- [34]
You have to use Avogadro's number (6.02x10^23 molecules/mole) to find the number of moles each reactant starts off with.
moles of Fe and O₂:
12 atoms/(6.02x10^23 atoms/mole)=1.99x10^-23 mol Fe
6 molecules/(6.02x10^23 molecules/mole)=9.967x10^-24 mol <span>O₂
</span>Then you find the limiting reagent by finding how much product each given amount of reactant can make.  Which ever one produces the least amount of product is the limiting reagent.
amount of Fe₂O₃ produced:
<span>(1.99x10^-23 mol Fe)x(2mol/4mol)= 9.967x10^-24mol Fe</span>₂O₃<span> 
</span>(9.967x10^-24 mol O₂)x(2mol/3mol)= 6.645x10^-24 mol Fe₂O₃<span> 
</span>since oxygen produces the leas amount of product, oxygen is the limiting reagent.  since we know that oxygen is the limiting reagent we can use the amount of product formed with oxygen to find the amount of iron used.
6.645x10^-24 mol Fe₂O₃x(4mol/2mol)=1.329x10^-23 mol Fe consumed
<span> find the amount left over by subtracting the original amount of Fe by the amount consumed in the reaction.
</span>1.993x10^-23-1.329x10^-23= 6.645x10^-23mol Fe left
 find the number of atoms by multiplying that by Avogadro's number.
<span>(6.645x10^-23mol)x(6.02x10^23 atoms/mol)=4 atoms
</span>therefore 4 atoms of Fe will be left over after the reaction happens.

I hope this helps.
8 0
3 years ago
Read 2 more answers
What is an important part of the endocrine system that acts as a chemical messenger?
olga_2 [115]
Hormones ! hope i helped :)
5 0
3 years ago
This type of chemist understands the structure of living systems and, in turn, their
julia-pushkina [17]

A Biochemist is a type of chemist understands the structure of living systems and, in turn, their functions and ways to control them.

<h3>What is the chemistry of living systems called?</h3>

The chemistry of living system is known as Biochemistry.

Biochemistry is a study of the chemical changes that occur in living organisms.

Scientists that study biochemistry are called Biochemists.

Biochemistry studies the structure and function of biological molecules such as carbohydrates, lipids, proteins, e.t.c., as well the chemical reaction they undergo.

Biochemistry also studies the energy changes that occur in living systems.

In conclusion, the chemistry of living systems is called Biochemistry.

Learn more about biochemistry at: brainly.com/question/12273783

#SPJ1

3 0
2 years ago
What is the percentage composition when 10g of magnesium combines with 4g of nitrogen?
Ivanshal [37]
The %  composition   when 10g of magnesium combine   with  4g of   nitrogen  is  71.43%   magnesium   and  28.57 %  nitrogen

               calculation

%
  composition  =  mass  of an element  / total mass  x100
mass  of magnesium = 10 g
mass of nitrogen  =  4g

calculate  the  total  mass  used

=  10g of  Magnesium  + 4 g of  nitrogen = 14 grams

%   composition for  magnesium  is therefore  =  10/14  x100 = 71.43 %

%  
composition  for  nitrogen  is therefore = 4 /14  x100  =   28.57 %
7 0
3 years ago
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