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atroni [7]
3 years ago
7

A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

te. (a) Write the molecular equation for the reaction. (b) What is the limiting reactant in the reaction? (c) How many grams of precipitate potentially form?
Chemistry
1 answer:
mestny [16]3 years ago
4 0

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

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Mrac [35]

Answer:

Mass of hydrogen gas evolved is 0.0749 grams.

Explanation:

Total pressure of the gases = p = 758 mmHg

Vapor pressure of water = 23.78 mmHg

Pressure of hydrogen gas ,P =  p - 23.78 mmHg = 758 mmHg - 23.78 mmHg

P = 734.22 mmHg = \frac{734.22}{760} atm=0.966 atm

Temperature  of of hydrogen gas ,T= 25°C =298.15 K

Volume of hydrogen gas = V = 0.949 L

Moles of hydrogen gas =n

PV = nRT (Ideal gas equation )

n=\frac{PV}{RT}=\frac{0.966 atm\times 0.949 L}{0.0821 atm L/mol K\times 298.15 K}

n = 0.03745 mol

Moles of hydrogen gas = 0.03745 mol

Mass of  0.03745 moles of hydrogen gas  = 0.03745 mol × 2 g/mol = 0.0749 g

Mass of hydrogen gas evolved is 0.0749 grams.

3 0
3 years ago
Hydrogen gas at a pressure of 740. mmHg has a volume of 2.00 L at a temperature of 25.0°C. What is the temperature of this gas a
nikdorinn [45]

Answer:

The final temperature of hydrogen gas is  537.63 K.

Explanation:

Given data:

Initial volume = 2.00 L

Initial pressure = 740 mmHg (740/760 = 0.97 atm)

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

Final temperature =?

Final volume = 3.50 L

Final pressure = standard = 1 atm

Formula:  

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂

T₂  =  P₂V₂T₁  / P₁V₁

T₂ = 1 atm × 3.5 L × 298 K / 0.97 atm × 2.00 L  

T₂ = 1043  atm .L. K / 1.94 atm. L

T₂ = 537.63 K

7 0
3 years ago
How is it possible for there to be more than one kind of atom of the same element?
tiny-mole [99]

They are called isotopes.

Isotopes have the same number of electrons and protons in their unionized state. They differ in the number of neutrons. The first and simplest example is hydrogen.


The most common hydrogen has

1 proton

1 electron and

0 neutrons


It has 2 cousins

1 proton

1 electron

1 neutron

And

1 proton

1 electron

2 neutrons.


Most elements have some differences in the number of neutrons present in their nuclei. Cesium and Xenon have the most number of isotopes. Each has 36. You wonder how the atoms are held together.


6 0
3 years ago
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kolezko [41]

Answer:

The types of Asexual reproduction include:  fission, fragmentation, budding, vegetative reproduction, spore formation and agamogenesis.

5 0
3 years ago
Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. When samples of these were decomposed the sulfur dioxide produce
Zinaida [17]

Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

Explanation : Given,

Mass of oxygen in sulfur dioxide = 3.49 g

Mass of sulfur in sulfur dioxide = 3.50 g

Mass of oxygen in sulfur trioxide = 9.00 g

Mass of sulfur in sulfur trioxide = 6.00 g

Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.

Mass of oxygen per gram of sulfur for sulfur dioxide = \frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}

Mass of oxygen per gram of sulfur for sulfur dioxide = \frac{3.49}{3.50}=0.997g

and,

Mass of oxygen per gram of sulfur for sulfur trioxide = \frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}

Mass of oxygen per gram of sulfur for sulfur trioxide = \frac{9.00}{6.00}=1.5g

Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

8 0
3 years ago
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