There are many ___ about ghost ships &sea monsters

What are compounds that react with acids and bases to make certain colors?

finlep [7]

Those would be called indicators , simply google them and you should find many. I'll leave some examples "bromophenol blue" "Methyl red" and "phenol red".

4 0

3 years ago

Jonathan claims that since only humans can read, it must be an inherited trait, passed from human parents to their children. Is

Crank

Jonathan is not correct because it has to be a trait that is learned.

So in that case Jonathan has to say that this is not an inherited trait it is learned by most people.

5 0

3 years ago

How many atoms are in 68.44 grams

creativ13 [48]

Correct answer There are 68 grams are in atoms

3 0

3 years ago

How many atoms are there in 8.88 g Si?

Mariana [72]

Answer:

$\boxed {\boxed {\sf 1.90 \times 10^{23} \ atoms \ Si}}$

Explanation:

We are asked to find how many atoms are in 8.88 grams of silicon.

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

First, we convert grams to moles. We use the molar mass or the mass of 1 mole of a substance. These values are found on the Periodic Table as they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up silicon's molar mass.

Si: 28.085 g/mol

We will convert using dimensional analysis. Set up a conversion factor with the molar mass.

$\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

We are converting 8.88 grams of silicon to moles, so we multiply by this value.

$8.88 \ g \ Si *\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

Flip the fraction so the units of grams of silicon cancel.

$8.88 \ g \ Si *\frac{1 \ mol \ Si} { 28.085 \ g \ Si}$

$8.88 *\frac{1 \ mol \ Si} { 28.085 }$

$\frac {8.88} { 28.085 } \ mol \ Si$

$0.316183015845 \ mol \ Si$

<h3>2. Moles to Atoms </h3>

Next, we convert moles to atoms. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of silicon.

Set up another conversion factor.

$\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

Multiply by the number of moles we calculated.

$0.316183015845\ mol \ Si *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

The units of moles of silicon cancel.

$0.316183015845 * \frac {6.022 \times 10^{23} \ atoms \ Si}{1}$

$0.316183015845 * {{6.022 \times 10^{23} \ atoms \ Si}$

$1.90405412 \times 10^{23} \ atoms \ Si$

<h3>3. Significant Figures</h3>

The original measurement of 8.88 grams has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 0 in the hundredth place.

$1.90 \times 10^{23} \ atoms \ Si$

<u>8.88 grams of silicon contains 1.90 ×10²³ atoms of silicon.</u>

6 0

3 years ago

How is a coefficient used to balance an equation

ludmilkaskok [199]

Let's start to understand this question by a simple combustion reaction involving oxidation of Ethane in the presence of Oxygen. When Ethane is burned in the presence of Oxygen it produces Carbon Dioxide and Water respectively. Therefore, the equation is as,

C₂H₆ + O₂ → CO₂ + H₂O

Above reaction shows the reaction and the equation is unbalanced. Balancing chemical equation is important because according to law of conservation of mass, mass can neither be created nor destroyed. Hence, we should balance the number of elements on both side.

LHS RHS

Carbon Atoms 2 1

Hydrogen Atoms 6 2

Oxygen Atoms 2 3

It means this equation is not obeying the law. Now, how to balance? One way is as follow,

C₂H₆ + O₃ → C₂O₂ + H₆O

LHS RHS

Carbon Atoms 2 2

Hydrogen Atoms 6 6

Oxygen Atoms 3 3

We have balanced the equation by changing the subscripts. But, we have messed up the chemical composition of compounds and molecules like Oxygen is converted into Ozone.

Therefore, we will change the coefficients (moles) to balance the equation as,

C₂H₆ + 7/2 O₂ → 2 CO₂ + 3 H₂O

LHS RHS

Carbon Atoms 2 2

Hydrogen Atoms 6 6

Oxygen Atoms 7 7

Now, by changing the coefficients we have balanced the equation without disturbing the chemical composition of compounds and molecules.

3 0

3 years ago