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Umnica [9.8K]
3 years ago
12

Does pressure decrease when a gas is removed

Chemistry
1 answer:
vovangra [49]3 years ago
4 0

yes

Explanation:

the volume also decrease. as particles are removed from the space the gas is in, there is a decrease in the number of collisions.

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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 an atomic radius?
Vikki [24]

Explanation:

The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons

8 0
3 years ago
Please help!!!!!<br> Please help!!!!!
lesantik [10]

Answer:

1.B

2.A

3. B

Explanation:

1. A chemical bond is the physical phenomenon of chemical substances being held together by attraction of atoms to each other through sharing

2. so 2 is electrons of one atom are transferred permanently to another atom.meaning the answer is A

3. is When atoms combine by forming covalent bonds, the resulting collection of atoms is called a molecule. We can therefore say that a molecule is the simplest unit of a covalent compound.

3 0
3 years ago
Read 2 more answers
What is it called when evaporation takes place beneath the surface of a liquid
Tems11 [23]
It is called boiling when evaporation takes place beneath the surface of a liquid. The evaporation is a kind of vaporization only takes place at the surface of a liquid. 
5 0
3 years ago
The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo
notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0
3 years ago
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