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3 years ago

1. What is the molarity of a solution which contains 0.256 mol of a substance dissolved in 143 mL of solution.

Chemistry

1 answer:

LenaWriter [7]3 years ago

4 0

Answer:

The equation for molarity is moles/liter for the first question you would do 0.256/0.143 liters to get 1.790 mol/L

Explanation:

The second problem you would do need to find the moles of NaCl which you would do by doing 4.89 g/58.44g/mol= 0.08367 then do 0.08367/0.600= 0.139 mol/L

The third problem would be the same steps as the second one.

The fourth problem would be (0.460M)(5.50L)= 2.53 moles

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Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$

From the reaction, we conclude that

As, 1 mole of $KI$ react to give 1 mole of $AgI$

So, 0.0178 moles of $KI$ react to give 0.0178 moles of $AgI$

Thus,

Moles of AgI = Moles of $I^-$ anion = Moles of $Ag^+$ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

$\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}$

$\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M$

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0

3 years ago

zloy xaker [14]

Answer:

1. They lack a nucleus

Following me..

5 0

3 years ago

Calculate the mols for 4.2g of a Mg

hjlf

Answer:

0.175mol

Explanation:

Mole of a substance can be calculated using the formula as follows:

number of moles (n) = mass (m) ÷ molar mass (MM)

According to this question, there are 4.2g of Magnesium (Mg).

Molar mass of Magnesium = 24g/mol, hence, the number of moles of 4.2g of Mg is as follows:

n = 4.2g ÷ 24g/mol

n = 0.175mol

8 0

3 years ago

Nitrogen effuses through a pinhole 1.7 times as fast as another gaseous element under the same conditions. Estimate the other el

marysya [2.9K]

Answer:

80.92, Krypton

Explanation:

What is effusion?

• It is a process where gas escapes through a pinhole (a very small hole) into a region of low pressure or vacuum

Graham's law of effusion of gas

• states that at a given constant temperature and pressure, the rate of effusion of gases is inversely proportional to the square root of their molar masses

$\boxed{ \frac{Rate_1}{Rate_2} = \sqrt{ \frac{M_2}{M_1} } }$

Calculations

Nitrogen exist as N₂ at room temperature, thus its molar mass is 2(14)= 28.

Let the rate and molar mass of unknown gas be Rate₂ and M₂ respectively.

Since N₂ effuses 1.7 times as fast as the unknown gas,

Rate₁= 1.7(Rate₂)

$\frac{Rate_1}{Rate_2} = 1.7$

$1. 7 = \sqrt{ \frac{M_2}{28} }$

Square both sides:

$2.89 = \frac{M_2}{28}$

Multiply both sides by 28:

2.89(28)= M₂

M₂= 80.92

Identity of gas

The molar mass of 80.92 lies between Bromine and Krypton. However since Bromine exist as Br₂, the value of it's molar mass would be 159.8 instead. Hence, Bromine is eliminated.

If the gas is a diatomic element, the atomic weight is 80.92 ÷2= 40.46. Thus, we are now considering if Argon could be its identity. However, Argon is a noble gas and will not exist as a diatomic element. Argon is therefore eliminated too.

Thus based on the above reasoning, its probable identity is Krypton.

7 0

3 years ago

The unheated Gas in the above system has a volume of 20.0 L at a temperature of 25.0 C and a pressure of 1.00 atm. The gas is he

kipiarov [429]

Answer:

1.25 atm.

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Initial volume (V1) = 20L

Initial temperature (T1) = 25°C

Initial pressure = 1 atm

Final temperature (T2) = 100°C

Final volume (V2) = constant i.e remain the same

Final pressure (P2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

Temperature (Kelvin) = temperature (celsius) + 273

Initial temperature (T1) = 25°C = 25°C + 273 = 298K

Final temperature (T2) = 100°C = 100°C + 273 = 373K

Step 3:

Determination of the final pressure of the gas. This is illustrated below:

Since the volume is constant, the following equation, P1/T1 = P2/T2 will be used to obtain the final pressure of gas as follow:

P1/T1 = P2/T2

Initial temperature (T1) = 298k

Initial pressure = 1 atm

Final temperature (T2) = 373K

Final pressure (P2) =?

P1/T1 = P2/T2

1/298 = P2 /373

Cross multiply to express in linear form

298 x P2 = 1 x 373

Divide both side by 298

P2 = 373/298

P2 = 1.25 atm.

Therefore, the pressure of the heated gas is 1.25 atm.

3 0

4 years ago