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kaheart [24]
3 years ago
14

At a certain temperature and pressure, 0.20 mol of carbon dioxide has a volume of 3.1 L. A 3.1-L sample of hydrogen at the same

temperature and pressure ____.
Chemistry
1 answer:
Misha Larkins [42]3 years ago
4 0
Answer:
            <span>A 3.1 L sample of hydrogen at the same temperature and pressure contains the same number of molecules.

Explanation:
                   As we know ;"1 mole of any gas at STP occupies exactly 22.4 L of volume". Therefore, CO</span>₂ and H₂ gas having same number of moles (0.20 moles) will occupy exactly the same volume.

Now, converting moles into number of molecules,

# of Molecules  =  Moles × 6.022 × 10²³

Putting value of mole,

# of Molecules  =  0.2 mol × 6.022 × 10²³ molecules.mol⁻¹

# of Molecules  =  1.20 × 10²³ molecules

Hence, both gases will contain 1.20 × 10²³ molecules.
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17) A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with what volume? HI
galina1969 [7]

Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

Explanation:

Given: M_{1} = 0.20 M,      V_{1} = 15.0 mL

M_{2} = 0.10 M,            V_{2} = ?

Formula used is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula s follows.

M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL

Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

4 0
2 years ago
Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g
Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

Volume of base,

= 30 ml

Molarity of base,

= 0.05 m

Molar mass of acid,

= 400 g/mol

As we know,

⇒  Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}

On substituting the values, we get

⇒           0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}

⇒  Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}

⇒                                             =1.5\times 10^{-3}  

hence,

⇒  Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}

On substituting the values, we get

⇒  1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}

⇒  Mass \ of \ acid=1.5\times 10^{-3}\times 400

⇒                         =0.60 \ g

8 0
3 years ago
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Aloiza [94]
5.00 x  1011/s = 5.05500kilohertz
7 0
2 years ago
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Formation of water: 2H2 + 1 O2 --&gt; 2H2O
noname [10]

Answer:

2.2 moles H2O

Explanation:

35g O_2 \mbox{ \cdot }\frac{1mol}{32g/mol} \mbox{ \cdot }\frac{2mol H_2O}{1mol O_2}= 2.1875, which rounds to about 2.2

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