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3 years ago

Why is Nertic zone particularly rich in living things?

Chemistry

1 answer:

Leya [2.2K]3 years ago

8 0

Answer:

It is rich in organisms because sunlight passes through its shallow water enabling photosynthesis to occur.

Explanation:

I hope that helped!!

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When trying to clean muddy, dirty river water, which method would work

Llana [10]

When trying to clean muddy, dirty river water, FILTRATION would work best

6 0

3 years ago

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Scrat [10]

Answer:

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4 years ago

How many molecules are in 3.6 grams of NaCl? Question options:

Levart [38]

Answer:

$\boxed {\boxed {\sf 3.7 * 10^{22} \ molecules \ NaCl}}$

Explanation:

We are asked to find how many molecules are in 3.6 grams of sodium chloride.

<h3>1. Convert Grams to Moles </h3>

First, we convert grams to moles using the molar mass. These values are equivalent to atomic masses on the Periodic Table, but the units are grams per moles instead of atomic mass units. Look up the molar masses of the individual elements: sodium and chlorine.

Na: 22.9897693 g/mol
Cl: 35.45 g/mol

There are no subscripts in the chemical formula (NaCl), so we simply add the 2 molar masses.

NaCl: 22.9897693 + 35.45 = 58.4397693 g/mol

Now we will convert using dimensional analysis. First, set up a ratio using the molar mass.

$\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}$

We are converting 3.6 grams to moles, so we must multiply the ratio by this value.

$3.6 \ g \ NaCl *\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}$

Flip the ratio so the units of grams of sodium chloride cancel.

$3.6 \ g \ NaCl *\frac { 1 \ mol \ NaCl}{58.4397693 \ g \ NaCl}$

$3.6 *\frac { 1 \ mol \ NaCl}{58.4397693}$

$\frac { 3.6}{58.4397693} \ mol \ NaCl$

$0.06160188589 \ mol \ NaCl$

<h3>2. Convert Moles to Molecules </h3>

Next, we convert moles to molecules using Avogadro's Number. This is 6.022 × 10²³ and it tells us the number of particles (atoms, molecules, formula units, etc). In this case, the particles are molecules of sodium chloride. Let's set up another ratio.

$\frac {6.022 \times 10^{23} \ molecules \ NaCl}{ 1 \ mol \ NaCl}$

Multiply by the number of moles we calculated.

$0.06160188589 \ mol \ NaCl * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 \ mol \ NaCl}$

The units of moles of sodium chloride cancel.

$0.06160188589 * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 }$

$3.70966557*10^{22} \ molecules \ NaCl$

<h3>3. Round </h3>

The original measurement of grams (3.6) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredth place tells us to leave the 7 in the tenth place.

$3.7 * 10^{22} \ molecules \ NaCl$

There are $3.7 * 10^{22} \ molecules \ NaCl$ in 3.6 grams and the correct answer is choice D.

5 0

3 years ago

How many numbers of atoms there are in 1.14 mol of (SO3)

Pachacha [2.7K]

called the Avogadro number

N(A)= 6.02 x 10^23 mol^-1

1 mole of SO3 will contain 6.02 x 10^23 mol^-1 of SO3 molecules.

thus, 1.14moles will contain;

= 1.14mol x [3mol O/1mol SO3] x [6.02 x 10^23

atoms O/1mol O]

= 2.05884 x 10^24 oxygen atoms

= 1.14mol x [1mol S/1mol SO3] x [6.02 x 10^23

atoms O/1mol O]

= 6.8628 x 10^23 sulfur atoms

hope this helps:-)

8 0

3 years ago

2. Suppose 13.7 g of C2H2 reacts with 18.5 g O2 according to the reaction below. C2H2(g) + O2(g) → CO2(g) + H2O(ℓ) a. What is th

max2010maxim [7]

Answer:

Mass of CO₂ produced = 20.328 g

Oxygen is limiting reagent.

Explanation:

Given data:

Mass of C₂H₂ = 13.7 g

Mass of O₂ = 18.5 g

Mass of CO₂ produced = ?

What is limiting reagent = ?

Solution:

Chemical equation:

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

Number of moles of C₂H₂:

Number of moles = mass /molar mass

Number of moles = 13.7 g/ 26.04 g/mol

Number of moles = 0.526 mol

Number of moles of O₂:

Number of moles = mass /molar mass

Number of moles = 18.5 g/ 32 g/mol

Number of moles = 0.578 mol

Now we will compare the moles of CO₂ with C₂H₂ and O₂

C₂H₂ : CO₂

2 : 4

0.526 : 4/2×0.526 = 1.052

O₂ : CO₂

5 : 4

0.578 : 4/5×0.578 = 0.462

The number of moles of CO₂ produced by O₂ are less thus oxygen will be limiting reactant.

Mass of CO₂ produced:

Mass = number of moles × molar mass

Mass = 0.462 mol × 44 g/mol

Mass = 20.328 g

7 0

3 years ago