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lukranit [14]
2 years ago
9

Each response gives a pair of solutions. which pair of solutions should conduct electrical current equally well? 1. 0.10 m nh3 a

nd 0.10 m nh4cl 2. 0.10 m nacl and 0.10 m naclo4 3. 0.10 m nano3 and 0.10 m hno2 4. 0.10 mhttps://yt3.ggpht.com:443
Chemistry
1 answer:
Crazy boy [7]2 years ago
3 0

Answer : The correct answer is option 2 : 0.10 M NaCl and 0.10 M NaClO₄

Explanation :

Solutions are classified into 3 categories.

1) Strong electrolytes : These are the solutions that dissociate completely forming ions . They are good conductors of electricity.

Example : All strong acids, bases and the salts made by strong acid/base are strong electrolytes .

2) Weak electrolytes : These substances do not dissociate completely thereby forming fewer ions. They are weak conductors of electricity.

3) Non electrolytes : These are the substances that do not dissociate at all. They do not form ions in aqueous medium. They are bad conductors of electricity.

Let us take a look at the given options and find out what type of solution do we have .

Option 1 : NH₃ is a weak electrolyte whereas NH₄Cl is a strong electrolyte bcause NH₄Cl is made by combination of NH₃ and HCl ( HCl is a strong acid)

Therefore NH₃ would carry electricity less efficiently than NH₄Cl.

Option 2 : Both NaCl and NaClO₄ are strong electrolytes. Therefore they will conduct electrical current equally well

Option 3 : NaNO₃ is a strong electrolyte but HNO₂ is a weak electrolyte. Therefore they will not carry the current equally

Therefore the correct option is option 2

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On a cool morning (12"C), a balloon is filled with 1.5 L of helium. By mid afternoon, the temperature has soared to 32°C. What i
ddd [48]

Answer:

1.6 L

Explanation:

Using Charle's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 1.5 L

V₂ = ?

T₁ = 12 °C

T₂ = 32 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (12 + 273.15) K = 285.15 K  

T₂ = (32 + 273.15) K = 305.15 K  

Using above equation as:

\frac{1.5}{285.15}=\frac{V_2}{305.15}

V_2=\frac{1.5\cdot \:305.15}{285.15}

New volume = 1.6 L

8 0
2 years ago
Dr. Franck cuts a bar of pure gold into smaller and smaller pieces. Will this action change the element that makes up the bar? E
Anvisha [2.4K]

Answer:

i believe this is a chemical or physical question? well your answer to that is no the element does not change because the gold is still gold it is still physical because you have just cut it into piece it is still gold

Explanation:

lmk if it was helpful :/

7 0
2 years ago
How many moles of CaC2 are needed to react completely with 45.0g H2O?
sertanlavr [38]

Answer:

In order to react with 45 g of water 1.25 moles of CaC₂ are required.

Explanation:

Given data:

Moles of CaC₂ needed = ?

Mass of water = 45.0 g

Solution:

Chemical equation:

CaC₂ + 2H₂O      →      C₂H₂ + Ca(OH)₂

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 45 g/ 18 g/mol

Number of moles = 2.5 mol

Now we will compare the moles of water and CaC₂ from balance chemical equation:

                     H₂O             :             CaC₂

                       2                :                1

                     2.5              :             1/2×2.5 =1.25 mol

In order to react with 45 g of water 1.25 moles of CaC₂ are required.

6 0
2 years ago
What is the molarity of a solution that contains 2.35 g of nh3 in 0.0500 l of solution?
tankabanditka [31]
The molarity is the number of moles in 1 L of the solution. 
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol 
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
3 0
1 year ago
Read 2 more answers
A chemist uses 0.25 L of 2.00 M H2SO4 to completely neutralize a 2.00 L of solution of NaOH. The balanced chemical equation of t
dalvyx [7]
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

c₁=2.00 mol/L
v₁=0.25 L
v₂=2.00 L
c₂-?

n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
n(NaOH)=2n(H₂SO₄)

c₂v₂=2c₁v₁

c₂=2c₁v₁/v₂

c₂=2*2.00*0.25/2.00=0.5 mol/L

0.5 M NaOH


4 0
2 years ago
Read 2 more answers
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