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KengaRu [80]
4 years ago
11

Something that does not have the ability to react is considered

Chemistry
1 answer:
nasty-shy [4]4 years ago
4 0
Something that does not have the ability to react is considered inert. <span> In chemistry, the term </span>inert<span> is used to describe a substance that is not </span>chemically<span> reactive. Hope this answers the question. Have a nice day.</span>
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What is the density of 2 mol of (He) at STP? (Show work and explain, please!!!)​
Svetradugi [14.3K]

Answer:

d = 1.8 × 10⁻⁴ kg/L

Explanation:

Given data:

Moles of helium = 2 mol

Temperature and pressure = standard

Density = ?

Solution:

PV = nRT

V = nRT/ P

V = 2 mol × 0.0821 atm. L/ mol. K × 273 K / 1 atm

V = 44.8 L

Mass of helium:

Mass = number of moles  × molar mass

Mass = 2 mol × 4 g/mol

Mass = 8 g

Mass = 0.008 kg

Density;

Density = mass/ volume

d = 0.008 kg/ 44.8 L

d = 1.8 × 10⁻⁴ kg/L

7 0
4 years ago
What are the two flasks for measuring the liquid volume?
aleksandrvk [35]

Answer:

Liquid volume is usually measured using either a graduated cylinder or a buret.

7 0
3 years ago
THIS IS A SCIENCE QUESTION
MissTica

Answer:

A

Explanation:

Hot air is less dense therefore it rises. Cold air is more closely packed therefore it is more dense. Less dense things rise while more dense things sink.

4 0
3 years ago
Read 2 more answers
Hello
nordsb [41]

Answer:

The ratio of the number of atoms of gold and silver in the ornament is 197 : 1080.

Explanation:

Let the mass of silver ornament be x.

Mass of gold polished on an ornament = 1% of x= 0.1 x

Moles of silver =\frac{x}{108 g/mol}

Number of atoms = Moles\times N_A

Where : N_A = Avogadro number

Moles of gold =\frac{0.1x}{197 g/mol}

Silver atoms =\frac{x}{108 g/mol}\times N_A

Gold atoms =\frac{0.1x}{197 g/mol}\times N_A

The ratio of the number of atoms of gold and silver in the ornament:

=\frac{\frac{0.1x}{197 g/mol}\times N_A}{\frac{x}{108 g/mol}\times N_A}

= 197 : 1080

3 0
3 years ago
Calculate the solubility (in g/L) of CaSO 4 ( s ) CaSO4(s) in 0.400 M Na 2 SO 4 ( aq ) at 25 ° C 0.400 M Na2SO4(aq) at 25°C. The
Allushta [10]

Explanation:

Ionization equation for CaSO_{4} is as follows.

     CaSO_{4} \rightarrow Ca^{2+} + SO^{2-}_{4}

        s              s           s

Now, the expression for the solubility product is as follows.

          K_{sp} = [Ca^{2+}][SO^{2-}_{4}]

                = s \times s

                = s^{2}

As the concentration of Na_{2}SO_{4} is given as 0.4 M.

So,  [Na_{2}SO_{4}] = [SO^{2-}_{4}] = 0.4 M

Putting the given values as follows.

           K_{sp} = [Ca^{2+}][SO^{2-}_{4}]

     4.93 \times 10^{-5} = [Ca^{2+}] \times 0.4

              [Ca^{2+}] = 12.325 \times 10^{-5}[/tex]

Hence, the solubility of CaSO_{4} in Na_{2}SO_{4} is 12.325 \times 10^{-5}.

Therefore, solubility of CaSO_{4} in g/ml as follows.

        12.325 \times 10^{-5} \times 136 g/mol

           = 0.0167 g/L

Thus, we can conclude that solubility of CaSO_{4} is 0.0167 g/L.

4 0
3 years ago
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