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3 years ago

What is the cation and anion of NaOH

Chemistry

2 answers:

mezya [45]3 years ago

7 0

Answer: Na (sodium ) is the cation, OH is the anion.

Explanation: Na (sodium) has an oxidation number of +1 since it is in group 1 alkali metals. Most metals are cations. For OH- (hydroxide) it has a negative 1 charge so it is the anion in the compound NaOH ( sodium hydroxide.

Wewaii [24]3 years ago

6 0

Answer:

naoh is called sodium hydroxide,

Explanation:

hope this helps

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When two gases or two liquids form a solution, the substance that is present in the largest amount is the?

skad [1K]

Is the solvent (Hope this helped)

4 0

3 years ago

How many moles of water as a gas can be formed 2.45 L

Alex17521 [72]

Answer:

0.11mole

Explanation:

Let us assume that the condition is at standard temperature and pressure(STP);

Given parameters:

Volume of water = 2.45L

Unknown:

Number of moles found in this volume of water = ?

Solution;

At STP;

Number of moles = $\frac{volume of gas}{22.4}$

Input the parameters and solve;

Number of moles of water = $\frac{2.45}{22.4}$ = 0.11mole

The number of moles of water found is 0.11mole

4 0

3 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

2 years ago

Can someone help with this please?

PtichkaEL [24]

A and b are related because if u look carefully at what its showing u that there both the same but what its telling u is not the same.

a and c are the same but the picture is different and also the way they describe it is different and what they want u to look at is that if you look at it closely then youll know the difference and how to find it as well

8 0

2 years ago

What is mass? What instrument is used to measure mass? What are the basic units of mass?

Oxana [17]

Mass is an amount of substance present in a body. mass is measure with the help of triple beam balance. the unit of mass are kilogram, gram etc.

6 0

2 years ago