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lesya [120]
2 years ago
5

How many liters of a 0.225 M solution of KI are needed to contain 0.935 moles of KI?

Chemistry
1 answer:
Katyanochek1 [597]2 years ago
6 0

Answer:

4.16L

Explanation:

From the question given, we obtained the following data:

Molarity = 0.225 M

Number of mole of KI = 0.935mole

Volume =?

Molarity = mole / Volume

Volume = mole /Molarity

Volume = 0.935/0.225

Volume = 4.16L

Therefore, 4.16L of KI is needed.

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What is the molar mass of a compound if 25.0 g of the compound dissolved in 750. Ml gives a molarity of 0.290 m?
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Answer:

115g/mol

Explanation:

To get the molar mass, we know that the it is equal to the mass divided by the number of moles. We have the mass but we do not have the number of moles.

We get this by working through the solution information. Firstly, we need to know the number of moles in 750ml for a molarity of 0.29m

Now, since 0.29 moles is present in 1000ml, x moles will be present in 750ml

The value of x is obtained as follows:

x = (750 * 0.29)/1000 = 0.2175 moles

Now since we have the number of moles, we can then obtain the molar mass.

Molar mass = mass/number of moles = 25.0g/0.2175 = 114.94 approximately 105g/mol

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2 years ago
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<span>Germanium is the element that has 32 protons in its nucleus.</span>
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How many milliliters of 0.02 M HCl are needed to react completely with 100 mL of 0.01 M NaOH?
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A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is
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Answer:

124.91mL

Explanation:

Given parameters:

P₁  = 1.08atm

V₁  = 250mL

T₁  = 24°C

P₂  = 2.25atm

T₂  = 37.2°C

V₂  = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

              \frac{P_{1} V_{1} }{T_{1} }   =  \frac{P_{2} V_{2} }{T_{2} }

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

        T₁  = 24°C,  T₁   = 24 + 273 = 297K

        T₂  = 37.2°C , T₂  = 37.2 + 273  = 310.2K

Input the variables and solve for V₂

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