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3 years ago

When balancing a chemical equation, what are the large numbers that we adjust?

Chemistry

1 answer:

Nata [24]3 years ago

6 0

Answer:

Coefficients

Explanation:

Chemical equations are first written as a skeleton equation, which includes how many atoms each element and compound has. Skeleton equations are not 'balanced' because the number of atoms of each element on the left side (reactants) is not equal to the right side (products).

To balance a chemical equation, you can write coefficients in front of single elements and compounds. The coefficient multiplies with each single element and with each element in the compound.

For example, in this skeleton equation:

H₂ + Cl₂ => HCl

Reactants: Products:

2 hydrogen 1 hydrogen

2 chlorine 1 chlorine

Write the coefficient 2 in the products.

H₂ + Cl₂ => 2HCl

Now both reactant and product sides have 2 chlorine and 2 hydrogen, so the equation is balanced.

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32 gm of O2 to mole of O2

lana [24]

Molar mass of O2:-

$\\ \rm\longmapsto 2(16u)=32g/mol$

Now

$\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \rm\longmapsto No\:of\;moles=\dfrac{32}{32}$

$\\ \rm\longmapsto No\:of\;moles=1mol$

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3 years ago

An inherited characteristic of an organism is called a ___________________.

Firlakuza [10]

Answer:

fitness. the ability of an organism to survive and reproduce in its environment. adaptation. the inherited characteristic that increases an organism's chance of survival.

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2 years ago

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How many moles of zinc are there in 0.890 g of zinc?

NNADVOKAT [17]

Answer:

There are

4.517

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atoms of Zn in 0.750 mols of Zn.

Explanation:

Since we know that there are

6.022

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atoms in every mole of a substance (Avogadro's Number), there are

6.022

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0.750

atoms of Zn in 0.750 mols of Zn.

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3 years ago

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Earnest Rutherford discovered that the nucleus is this

Nikitich [7]

Rutherford, through his gold-foil experiment, discovered that the nucleus is a small core of protons and neutrons making up most of the atom’s mass.

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3 years ago

Caffeine, a stimulant found in coffee and soda, hasthe mass percent composition: C. 49.48%, H, 5.19%. N. 28.85% 0. 16.48% The mo

borishaifa [10]

We have the next % composition:

C. 49.48%

H, 5.19%.

N. 28.85%

0. 16.48%

We assume 100 g of sample

1) As we have 100 g of sample of Caffeine, we calculate the mass of each element involved here.

C. 49.48 g

H, 5.19 g

N. 28.85 g

0. 16.48 g

2) We calculate the number of moles of each element (we need the mass per mole of each element)

For C) 12.01 g/mol

49.48 g x (1 mol/12.01 g) = 4.120 moles

For H) 1.007 g/mol

5.19 g x (1 mol/1.007 g) = 5.154 moles

For O) 15.99 g/mol

16.48 g x (1 mol/15.99 g) = 1.030 moles

For N) 14.00 g/mol

28.85 g x (1 mol/14.00 g) = 2.060 moles

3) We choose the smallest number from 2) and divide the rest of them by it.

For C) 4.120 moles/1.030 moles= 4

For H) 5.154 moles/1.030 moles= 5

For O) 1.030 moles/1.030 moles= 1

For N) 2.060 moles/1.030 moles= 2

4) The numbers in 3) represents the subindex from the empirical formula of caffeine:

$C_4H_5O_1N_2$

5) We calculate the molar mass of our empirical formula, 97.06 g/mol.

We already have the molar mass of the molecular formula, so we proceed like this:

n= the molar mass of the molecular formula/the molar mass of the empirical formula

n = 194.19 g/mol/97.06 g/mol = 2 approx.

We use "n" and we multiply our empirical formula by n = 2:

Therefore, our molecular formula:

$C_8H_{10}O_2N_4$

8 0

1 year ago