If you have 45.6mL of a liquid and its

Calculate the molarity of a solution with 233.772g sodium chloride dissolved in 2,000mL of water

LekaFEV [45]

Answer:

The molarity is 2M

Explanation:

First , we calculate the weight of 1 mol of NaCl:

Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol

58,5 g---1 mol NaCl

233,772 g--------x= (233,772 g x1 mol NaCl)/58,5 g= 4 mol NaCl

A solution molar--> moles of solute in 1 L of solution:

2 L-----4 mol NaCl

1L----x0( 1L x4mol NaCl)/4L =2moles NaCl---> 2 M

3 0

2 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Lera25 [3.4K]

Answer:

NH3

Explanation:

2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)

So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.

n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol

n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol

From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2

4 0

3 years ago

PLEASE HELP ME what would be the mass of 9.03 x 1021 molecules of hydrobromic acid (HBr)?

Nikitich [7]

We know 1 mole of any atom or molecules contains $6.033 \times 10^{23}$ atom or molecules.

1 mole of HBr i.e 81 gm/mol contains $6.033 \times 10^{23}$ atom or molecules.

So, mass of $9.03\times 10^{21}$ molecules is :

$m=\dfrac{81\times 9.03\times 10^{21} }{6.033 \times 10^{23}}\\\\m= 1.21\ gm$

Therefore, mass of $9.03\times 10^{21}$ molecules is 1.21 gm .

Hence, this is the required solution.

7 0

3 years ago

Acetylene gas, C2H2, can be produced by the reaction of calcium carbide and water. CaC2(s) + 2H2O(l) --> C2H2(g) + Ca(OH)2(aq

nekit [7.7K]

Answer:

1.0 L

Explanation:

Given that:-

Mass of $CaC_2$ = $2.54\ g$

Molar mass of $CaC_2$ = 64.099 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{2.54\ g}{64.099\ g/mol}$

$Moles_{CaC_2}= 0.0396\ mol$

According to the given reaction:-

$CaC_2_{(s)} + 2H_2O_{(l)}\rightarrow C_2H_2_{(g)} + Ca(OH)_2_{(aq)}$

1 mole of $CaC_2$ on reaction forms 1 mole of $C_2H_2$

0.0396 mole of $CaC_2$ on reaction forms 0.0396 mole of $C_2H_2$

Moles of $C_2H_2$ = 0.0396 moles

Considering ideal gas equation as:-

$PV=nRT$

where,

P = pressure of the gas = 742 mmHg

V = Volume of the gas = ?

T = Temperature of the gas = $26^oC=[26+273]K=299K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles = 0.0396 moles

Putting values in above equation, we get:

$742mmHg\times V=0.0396 mole\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\V=\frac{0.0396\times 62.3637\times 299}{742}\ L=1.0\ L$

1.0 L of acetylene can be produced from 2.54 g $CaC_2$ .

4 0

3 years ago

A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit

nexus9112 [7]

Answer:

pH = 3.70

Explanation:

Moles of aniline in solution are:

0.0500L × (0.0350mol / 1L) = 1.750x10⁻³ mol aniline

Aniline is in equilibrium with water, thus:

C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

HCl reacts with aniline thus:

HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing C₆H₅NH₃⁺. C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = 2.63x10⁻⁵ (1)

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ = [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

pH = 3.70

7 0

3 years ago