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xxMikexx [17]
4 years ago
15

How do you balance the following equation only using coefficients?

Chemistry
1 answer:
yKpoI14uk [10]4 years ago
3 0

The answer will be 6CO2 + 6H2O ——-> 1 C6H12O6 + 6O2

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DerKrebs [107]

two hydrogen atoms and one oxygen atom. hope this helped.

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3 years ago
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What is the mass of 9 atom(s) of copper in grams?
il63 [147K]
9.05 E-22 g. ~ 10.0 E-22 g Cu
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100 points! Formation of an Ion
RUDIKE [14]
\huge\bold {Answer}

Aluminum ions have a charge of +3. This means that the atoms have three more protons than electrons. In order to add electrons to aluminum ions, the purification process therefore requires a large amount of electricity.

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ion
atom that has a positive or negative charge because it lost or gained one or more electrons

chemical bond
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5 0
3 years ago
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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, q=mc\Delta T .

where, q is the heat energy, m is mass, c is specific heat and \Delta T is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is \frac{1.035cal}{g.^0C} .

\Delta T  for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

\Delta T  for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.


3 0
4 years ago
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How can so many different substances in the world be made of so few elements?
Mrrafil [7]
Matematically speaking, maybe because:
The number of substances = number of elements + number of different combinations of those elements
7 0
4 years ago
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