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Andre45 [30]
3 years ago
12

19) What is the molarity of a KOH solution if 200 ml of the solution contains 0.6 moles KOH?

Chemistry
2 answers:
zloy xaker [14]3 years ago
8 0

Answer: 3M

Explanation: Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula used :

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute KOH = 0.6 moles

V_s = volume of solution in ml= 200 ml

Now put all the given values in the formula of molarity, we get

Molarity=\frac{0.6moles\times 1000}{200ml}=3mole/L

Therefore, the molarity of solution will be 3M.

alexira [117]3 years ago
5 0

Answer:

3 moles

Explanation:

200 ml is 1/5 of a liter, so the answer is five times the number of moles present in the solution. 0.6 moles/0.2 liter = x moles/1.0 liter. Solving for x gives 0.2 x = 0.6 or x = 3.0 M.

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3 years ago
URGENT
Sergio [31]

T = 2T , P = 2P as given in the question

Volume = mass *density

Now apply ideal gas law

PV=nRT

V = nRT/P

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4 0
3 years ago
Consider the reaction below. Which species is(are) the Brønsted-Lowry acid(s)?
madreJ [45]

Answer:

HF is the acid

Explanation:

The Brønsted-Lowry theory defines the acids and bases in chemistry as follows:

An acid is the species that can donate a proton

A base can accept protons.

In the reaction:

HF(aq) + NH₃(aq) → NH₄⁺(aq) + F⁻(aq)

As you can see, HF can donate its proton to produce F⁻: HF is the acid

<em>In the same way, NH₃ is accepting a proton, NH₃ is the base.</em>

5 0
3 years ago
A person walks at 2 km/h for four hours. What is their displacement?
Masteriza [31]

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Explanation:

3 0
3 years ago
The molarity of an aqueous solution of hydroiodic acid, HI, is determined by titration with a 0.145 M potassium hydroxide, KOH,
hodyreva [135]

Answer:

0.133 M

Explanation:

The volume of the solution is given, so in order to find concentration, the number of moles must be found, since C = n/V.

The balanced reaction equation is:

HI + KOH ⇒ H₂O + KI

Thus, the moles of KOH added to neutralize all of the HI will be equal to the moles of HI that must have been present.

The amount of KOH that was added is calculated as follows.

n = CV = (0.145 mol/L)(45.7 mL) = 6.6265 mmol KOH = 6.6265 mmol HI

Since HI and KOH are related in a 1:1 molar ratio, the same amount of HI must have been present.

Finally, the concentration of HI is calculated:

C = n/V = (6.6265 mmol) / (50.0 mL) = 0.133 mol/L = 0.133 M

5 0
3 years ago
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