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2 years ago

How many silicon atoms are there in 28.09 g of Si?

Chemistry

2 answers:

irinina [24]2 years ago

8 0

Explanation:

28.09 g of silicon contains 6.02×1023 6.02 × 10 23 silicon atoms.

mixer [17]2 years ago

3 0

1 mole = 6.02.10²³ particles (atoms, moleculs, ions)

mole Si = mass : molar mass

mole Si = 28.09 : 28.09 = 1

number of atoms = 1 x 6.02.10²³ = 6.02.10²³ atoms

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6. What is the wavelength (in meters) of the electromagnetic carrier wave transmitted by The Sports

Vsevolod [243]

Answer:

wavelength = 2.81 m

Explanation:

Given that,

The frequency of the Sports Fan radio station, f = 106.4 MHz

We need to find the wavelength (in meters) of the electromagnetic carrier wave. The relation between frequency and wavelength is given by :

$c=f\lambda$

Where,

$\lambda$ is wavelength

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{106.4\times 10^6}\\\\\lambda=2.81\ m$

So, the wavelength of the electromagnetic carrier wave is 2.81 m.

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2 years ago

In what ways do scientists not study the history of Earth's climate?

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3 years ago

Which two species are most closely related

Eddi Din [679]

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3 years ago

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago

Which of these have the same number of particles as 1 mole of water H2O

Fed [463]

Answer:

It is equal to Avogadro's number (NA), namely 6.022 x1023. If we have one mole of water, then we know that it will have a mass of 2 grams (for 2 moles of H atoms) + 16 grams (for one mole O atom) = 18 grams.

Explanation:

The question is not very much clear.

If you are asking for molecules then 1 mole water= 6.023 * 10^23

If you are asking for atoms then 1 mole water= 6.023 * 10^23 * 3

If you are asking for particles then,

So, in your example you would have one mole of water molecules. If you dissociated those water molecules, than you would end up with 2 moles of hydrogen atoms, and one mole of oxygen atoms.

I hope that was helpful!

H=1 proton,1 electron

O=8 protons,8 neutrons and 8 electrons

total particles in one H2O molecule-28

total no. of particles in 1 mole of water- 6.023 * 10^23 * 28

8 0

2 years ago