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2 years ago

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Chemistry

2 answers:

xenn [34]2 years ago

8 0

Answer:

YES YES YES YES YES YES YES YES YES YES YES YES

Goryan [66]2 years ago

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Answer:

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Explanation:

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You might be interested in

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

What is the [OH-] of a substance that has a pH of 11?

Deffense [45]

Answer:

0.001 M OH-

Explanation:

[OH-] = 10^-pOH, so

pOH + pH = 14 and 14 - pH = pOH

14 - 11 = 3

[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-

6 0

2 years ago

What are some similarities and differences between physical and chemical weathering

mote1985 [20]

Similarities:
they both made sediment into soil
they both form the earth
they both made sediments have cracks

differences:
physical is reliant usually on contact with atmospheric condition

chemical transforms rocks into sediments while physical only breaks it down

chemical uses chemical reactions

(a brainliest would be appreciated)

3 0

3 years ago

Read 2 more answers

At a birthday pool party, the temperature is 28.50°C and the atmospheric pressure is 755.4 mmHg. One of the decoration helium ba

Iteru [2.4K]

Answer:

The volume of the balloon will be 5.11L

Explanation:

An excersise to solve with the Ideal Gases Law

First of all, let's convert the pressure in mmHg to atm

1 atm = 760 mmHg

760 mmHg ___ 1 atm

755.4 mmHg ____ (755.4 / 760) = 0.993 atm

922.3 mmHg ____ ( 922.3 / 760) = 1.214 atm

T° in K = 273 + °C

28.5 °C +273 = 301.5K

26.35°C + 273= 299.35K

P . V = n . R .T

First situation: 0.993atm . 6.25L = n . 0.082 . 301.5K

(0.993atm . 6.25L) / 0.082 . 301.5 = n

0.251 moles = n

Second situation:

1.214 atm . V = 0.251 moles . 0.082 . 301.5K

V = (0.251 moles . 0.082 . 301.5K) / 1.214 atm

V = 5.11L

7 0

3 years ago

Read 2 more answers

You are given a solid that is a mixture of na2so4 and k2so4.

Murljashka [212]

Here we have to calculate the amount of $SO_{4}^{2-}$ ion present in the sample.

In the sample solution 0.122g of $SO_{4}^{2-}$ ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ + $SO_{4}^{2-}$

(Na)₂SO₄=2Na⁺ + $SO_{4}^{2-}$

Thus, BaCl₂+ $SO_{4}^{2-}$ = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of $SO_{4}^{2-}$ ion is precipitated in this reaction.

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion ( $SO_{4}^{2-}$ ) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present. So in 1 g of BaSO₄ $\frac{96.06}{233.3}=0.411$ g of $SO_{4}^{2-}$ ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of $SO_{4}^{2-}$ ion is present.

5 0

3 years ago