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2 years ago

How much nano3 is needed to prepare 225 ml of a 1.55 m solution of nano3?

Chemistry

1 answer:

andrezito [222]2 years ago

5 0

Answer:

= 29.64 g NaNO3

Explanation:

Molarity is given by the formula;

Molarity = Moles/Volume in liters

Therefore;

Number of moles = Molarity × Volume in liters

= 1.55 M × 0.225 L

= 0.34875 moles NaNO3

Thus; 0.34875 moles of NaNO3 is needed equivalent to;

= 0.34875 moles × 84.99 g/mol

= 29.64 g

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1. Fill in the blanks. Example: A nitrogen atom takes on a 3- charge when it forms an anion and becomes nitride. a. A magnesium

Tema [17]

Answer: a. +2, cation and magnesium ion .

b. -1, anion, chloride

c. -2, anion, oxide

d. +1. cation , potassium ion

Explanation:

When an atom accepts an electron negative charge is created on atom and is called as anion.

When atom loses an electron positive charge is created on atom and is called as cation.

Magnesium (Mg) with atomic number of 12 has electronic configuration of 2,8,2 and thus it can lose 2 electrons to form $Mg^{2+}$ cation and becomes magnesium ion.

Chlorine (Cl) with atomic number of 17 has electronic configuration of 2,8,7 and thus it can gain 1 electron to form $Cl^{-}$ anion and becomes chloride.

Oxygen (O) with atomic number of 8 has electronic configuration of 2,6 and thus it can gain 2 electrons to form $O^{2-}$ anion and becomes oxide.

Potassium (K) with atomic number of 19 has electronic configuration of 2,8,8,1 and thus it can lose 1 electron to form $K^{+}$ cation and becomes potassium ion.

3 0

2 years ago

A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II) ion solution and another electrode c

Andre45 [30]

Answer: The standard potential of the cell is 0.77 V

Explanation:

We know that:

$E^o_{Ni^{2+}/Ni}=-0.25V\\E^o_{Cu^{+}/Cu}=0.52V$

The substance having highest positive $E^o$ reduction potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

Oxidation half reaction: $Ni(s)\rightarrow Ni^{2+}(aq)+2e^-$

Reduction half reaction: $Cu^{+}(aq)+e^-\rightarrow Cu(s)$ ( × 2)

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation follows:

$E^o_{cell}=0.52-(-0.25)=0.77V$

Hence, the standard potential of the cell is 0.77 V

5 0

2 years ago

3 NO2(g) + H2O(ℓ) −→

gulaghasi [49]

The balanced equation for the above reaction is as follows;
3NO₂ + H₂O --> 2HNO₃ + NO
stoichiometry of NO₂ to NO is 3:1
molar volume is where 1 mol of any gas occupies a volume of 22.4 L
volume of gas is directly proportional to number of moles of gas.
therefore stoichiometry can be applied for volume as well.
volume ratio of NO₂ to NO is 3:1
volume of NO₂ reacted - 854 L
therefore volume of NO formed - 854 L /3 = 285 L
volume of NO formed - 285 L

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2 years ago

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2 years ago

A 0.005 M potassium hydroxide solution is an example of what type of solution? *

poizon [28]

Answer:

Basic solution

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A solution containing 0.005 M potassium hydroxide has a pH of;

pOH= -log[0.005]

pOH= 2.3

But pH+pOH=14

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pH= 14-pOH

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2 years ago