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3 years ago

What is the representative particle for silicon

1 answer:

7 0

<span>The representative particle for silicon is atom of silicon.
</span>Representative particles can be atoms, molecules, formula units or ions. Representative particles depend on the nature of the substance. Silicon is chemical element made of atoms.
For sodium chloride (NaCl) for example, representative particles are ions.

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A sample of nitrogen containing 3.0×10^23 molecules has the same number of molecules as a sample containing

Daniel [21]

Answer:

D) 0.50 mole of Ne

Explanation:

Given data:

Number of molecules of nitrogen = 3.0×10²³ molecules

Which sample contain same number of molecules as nitrogen= ?

Solution:

A) 0.25 mole of O₂

1 mole = 6.022×10²³ molecules

0.25 mol × 6.022×10²³ molecules / 1 mol

1.51×10²³ molecules

B) 2.0 moles of He.

1 mole = 6.022×10²³ molecules

2.0 mol × 6.022×10²³ molecules / 1 mol

12.044×10²³ molecules

C) 1.0 moles of H₂

1 mole = 6.022×10²³ molecules

D) 0.50 mole of Ne

1 mole = 6.022×10²³ molecules

0.50 mol × 6.022×10²³ molecules / 1 mol

3.0×10²³ molecules

5 0

3 years ago

What is the purpose of an ionic bond?

Yuri [45]

Answer:

Ionic bonds are important because they allow the synthesis of specific organic compounds. Scientists can manipulate ionic properties and these interactions in order to form desired products. Covalent bonds are especially important since most carbon molecules interact primarily through covalent bonding

Explanation:

hope this helps!! have a great day!! (can u mark me brailyest plzz) :D

4 0

3 years ago

A company wants to put a fleet of low Earth satellites in orbit to create a worldwide telephone link. There is less of a time la

Marianna [84]

If your wanting to know less of the time:

70

250

8 0

3 years ago

You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho

timama [110]

Explanation:

As per the given data, at a higher temperature, at $24^{o}C$ , the solution will occupy a larger volume than at $20^{o}C$ .

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At $20^{o}C$ , density of water=0.9982071 g/ml

Therefore, $\frac{concentration}{density}$ will be calculated as follows.

= $\frac{C_{1}}{d_{1}}$

= $\frac{1.000 mol/L}{0.9982071 g/ml}$

= 1.0017961 mol/g

At $24^{o}C$ , density of water = 0.9972995 g/ml

Since, $\frac{concentration}{density}$ = $\frac{C_{2}}{d_{2}}$

= $\frac{C_{2}}{0.9972995}$

Also, $\frac{C_{1}}{d_{1}}$ = $\frac{C_{2}}{d_{2}}$

so, 1.0017961 mol/g = $\frac{C_{2}}{0.9972995}$

$C_{2} = 1.0017961 \times 0.9972995$

= 0.9990907 mol/L

Therefore, in 500 ml, concentration of $KNO_{3}$ present is calculated as follows.

$C_{2}$ = $\frac{concentration}{volume}$

0.9990907 mol/L = $\frac{concentration}{0.5 L}$

concentration = 0.49954537 mol

Hence, mass (m'') = $0.49954537 mol \times 101.1032 g/mol$ = 50.5056 g (as molar mass of $KNO_{3}$ = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

m = $m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}$

where, $d_{air}$ = density of air = 0.0012 g/ml

$d_{weight}$ = density of callibration weights = 8.0g/ml

d = density of weighed object

Hence, the true mass will be calculated as follows.

True mass(m) = $50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}$

true mass(m) = 50.5268 g

= 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of $KNO_{3}$ needs to be measured.

8 0

3 years ago

A container holds 500 mL of oxygen gas at a pressure of 1.3 atm and a temperature of 25 ˚C. If the temperature increases to 29 ˚

Ksivusya [100]

Answer:

The new volume of the gas is decreased, to 0.329 L

Explanation:

Step 1: Data given

Volume of oxygen = 500 mL = 0.500 L

Pressure = 1.3 atm

Temperature = 25.0 °C

Temperature increases to 29.0 °C

New pressure = 2.0 atm

Step 2: Calculate new volume

p1*V1/T1 = p2*V2/T2

⇒with p1 = the initial pressure = 1.3 atm

⇒with V1 = the initial volume = 0.500L

⇒with T1 = the initial temperature = 25 °C = 298 K

⇒with p2 = the new pressure = 2.0 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the new temperature = 29 °C = 302 K

(1.3 * 0.500)/298 = (2.0 * V2) /302

0.0021812 = 2.0V2 / 302

V2 = 0.329 L = 329 mL

The new volume of the gas is decreased, to 0.329 L

3 0

4 years ago