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3 years ago

Cu+2AgNO

Chemistry

1 answer:

baherus [9]3 years ago

6 0

Answer:

Mass of Ag produced = 64.6 g

Note: the question is, how many grams of Ag is produced from 19.0 g of Cu and 125 g of AgNO3

Explanation:

Equation of the reaction:

Cu + 2AgNO3 ---> 2Ag + Cu(NO3)2

From the equation above, 1 mole of Cu reacts with 2 moles of AgNO3 to produce 2 moles of Ag and 1 mole of Cu(NO3)2.

Molar mass of the reactants and products are; Cu = 63.5 g/mol, Ag = 108 g/mol, AgNO3 = 170 g/mol, Cu(NO3)2 = 187.5 g/mol

To determine, the limiting reactant;

63.5 g of Cu reacts with 170 * 2 g of AgNO3,

19 g of Cu will react with (340 * 19)/63.5 g of AgNO3 =101.7 g of AgNO3.

Since there are 125 g of AgNO3 available for reaction, it is in excess and Cu is the limiting reactant.

63.5 g of Cu reacts to produce 108 * 2 g of Ag,

19 g of Cu will react to produce (216 * 19)/63.5 g of Ag = 64.6 g of Ag.

Therefore mass of Ag produced = 64.6g

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If I initially have a gas at a pressure of 12 atm, volume of 23 liters, and temperature of 200 K, and then I raise the pressure

CaHeK987 [17]

Answer : The volume of gas will be 29.6 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 12 atm

$P_2$ = final pressure of gas = 14 atm

$V_1$ = initial volume of gas = 23 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 200K

$T_2$ = final temperature of gas = 300K

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{12atm\times 23L}{200K}=\frac{14\times V_2}{300K}$

$V_2=29.6L$

Therefore, the new volume of gas will be 29.6 L

5 0

3 years ago

Read 2 more answers

I RLLY NEED HELP. Use Image B (picture above) and calculate the density of a ring that has a mass of 32 grams. Read Page 11 "Mea

dimaraw [331]

Answer:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring = 8 g/mL

Explanation:

From the question given above, the following data were obtained:

Mass of ring = 32 g

Volume of water = 64 mL

Volume of water + ring = 68 mL

Density of ring =?

Next, we shall determine the volume of the ring. This can be obtained as follow:

Volume of water = 64 mL

Volume of water + ring = 68 mL

Volume of ring =?

Volume of ring= (Volume of water + ring) – (Volume of water)

Volume of ring = 68 – 64

Volume of ring = 4 mL

Finally, we shall determine the density of the ring. This can be obtained as follow:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring =?

Density = mass / volume

Density of ring = 32 / 4

Density of ring = 8 g/mL

8 0

3 years ago

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0

o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2) Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min 0 10 20 30 40 ∞

conc B/(mol/L) 0 0.089 0.153 0.200 0.230 0.312

For a first order reaction:

$K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})$

where :

K = proportionality constant or the rate constant for the specific reaction rate

t = time of reaction

$C_o$ = initial concentration at time t

$C _{\infty}$ = final concentration at time t

$C_t$ = concentration at time t

To start with the value of t when t = 10 mins

$K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})$

$K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})$

$K_1 =0.03358 \ min^{-1}$

$K_1 \simeq 0.034 \ min^{-1}$

When t = 20

$K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})$

$K_2= 0.05 \times \ In ( 1.9623)$

$K_2=0.03371 \ min^{-1}$

$K_2 \simeq 0.034 \ min^{-1}$

When t = 30

$K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})$

$K_3= 0.0333 \times \ In ( \dfrac{0.312}{0.112})$

$K_3= 0.0333 \times \ 1.0245$

$K_3 = 0.03412 \ min^{-1}$

$K_3 = 0.034 \ min^{-1}$

When t = 40

$K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})$

$K_4=0.025 \times \ In ( \dfrac{0.312}{0.082})$

$K_4=0.025 \times \ In ( 3.8048)$

$K_4=0.03340 \ min^{-1}$

We can see that at the different time rates, the rate constant of $k_1, k_2, k_3, and k_4$ all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0

3 years ago

A filament of a light bulb is made from a pure sample of tungsten.

GenaCL600 [577]

Answer:

The correct options are:

1) Pure chemical substance

2) Element.

Explanation:

each of the term is explained below:

1) Mixture: Since tungsten is an element in periodic table thus it is not a mixture.

2) False by definition of mixture.

3) false by definition of mixture.

4) Solution: Tungsten in a filament is in solid form hence solution is incorrect.

5) Since it is given in the question itself that the sample is pure hence option 'E' is correct as chemical substance is a pure substance that has a homogeneous composition and the sample in the question is given as pure.

6) Compound: By definition of compound it is formed by mixing 2 or more elements but since tungsten is an element that occurs independently in nature hence it is not a compound.

7) Tungsten is a element in the periodic table with atomic number 74.

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3 years ago

Which of these is a chemical property of a substance

Archy [21]

A chemical property is a characteristic of a substance that may be observed when it participates in a chemical reaction. Examples of chemical properties include flammability, toxicity, chemical stability, and heat of combustion

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3 years ago