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Snowcat [4.5K]
3 years ago
15

8.7g+15.43g+19g? 853.2L-627.443L? 5.47m+11m+87.300m?

Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0

the god ran down the street

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Explanation:

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What is the mass in grams of 0.280 mole sample of sodium hydroxide NaOH​
vivado [14]

Answer:

The mass of 0.280 mole sample of sodium hydroxide NaOH​ is 11.2 grams.

Explanation:

To know the mass in grams of 0.280 moles of sample of sodium hydroxide NaOH, you must know the molar mass of the compound, that is, the mass of one mole of a substance, which can be an element or a compound.

So you know:

  • Na: 23 g/mole
  • O: 16 g/mole
  • H: 1 g/mole

So, the molar mass of NaOH is:

NaOH= 23 g/mole + 16 g/mole+ 1 g/mole= 40 g/mole

Then the following rule of three can be applied: if in 1 mole of sodium hydroxide there are 40 grams, in 0.280 moles how much mass is there?

mass=\frac{0.28 moles*40 grams}{1 mole}

mass= 11.2 grams

<u><em>The mass of 0.280 mole sample of sodium hydroxide NaOH​ is 11.2 grams.</em></u>

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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

4 0
4 years ago
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