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3 years ago

How many particles would one mole of NaCl have in solution

Chemistry

1 answer:

galben [10]3 years ago

8 0

6.023 x 10^23 . Common salt . Total of 12.046 x 10^23 atoms

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kaheart [24]

Answer:

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Explanation:

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2 years ago

An aluminum bar was found to have a mass of 27g. Using water displacement, the volume was measured to be 10 ml. What is the dens

Pavlova-9 [17]

Answer:

2.7 g/mL:)

An aluminum bar was found to have a mass of 27g. Using water displacement, the volume was measured to be 10 ml. What is the density of the aluminum? Group of answer choices (27 g)/(10 ml) (10 ml )/(2.70 g) (270 g)/(10 ml) (10 ml )/(27 g)

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3 years ago

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

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3 years ago

I suck at this stuff. HELP

erastovalidia [21]

What do you need help with

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3 years ago

Based on the table, which do you think is larger – your direct water use or your indirect water use?

Pavlova-9 [17]

Answer:

your indirect water use.

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3 years ago

How many particles would one mole of NaCl have in solution​

How many particles would one mole of NaCl have in solution