All categories

3 years ago

Cu + AgNO3 → Ag + CuNO3

Chemistry

1 answer:

tia_tia [17]3 years ago

4 0

Answer:

26.0g of Ag

Explanation:

Equation of reaction,

Cu + AgNO3 → Ag + CuNO3

We need to find the mass of Ag produced in the reaction by comparing it with the mass of Cu reacting.

From the equation of reaction,

1 mole of Cu reacts with 1 mole of Ag

Number of mole = mass / molarmass

Molar mass of Cu = 63.55g

Molar mass of Ag = 107.87g

Number of moles = mass / molar mass

Mass = number of moles × molar mass

Mass of Cu = 1 × 63.55g

Mass of Cu = 63.55g

Similarly,

Mass of Ag = 1 × 107.87

Mass of Ag = 107.87g.

Now let's go back to the equation of reaction,

1 mole of Cu = 1 mole of Ag;

63.55g of Cu = 107.87g of Ag

15.32g of Cu = x g of Ag

X = (107.87 × 15.32) / 63.55

X = 1652.568 / 63.55

X = 26.00g

26.0g of Ag will be produced when 15.32g of Cu reacts with excess AgNO3

You might be interested in

What does the absolute magnitude of a star depend on?

Serggg [28]

C. Distance from Earth

The observed brightness of a star (the apparent magnitude) to an absolute magnitude, you need to know the distance [d], to the star. Alternatively, if we know the distance and the apparent magnitude [m] of a star, we can calculate its absolute magnitude [d].

*paraphrased from "COSMOS.com"

8 0

4 years ago

Anabelle has noticed that her strawberry plants haven't produced any strawberries yet this year.

marusya05 [52]

Answer:0.260

Explanation:

4 0

4 years ago

Where would the electrons in the 3d sublevel be found? Possibly in any shell only in the n = 3 energy level in energy levels &gt

postnew [5]

Answer:

In the n = 3 energy level

Explanation:

There's is no further explanation for this.

All the electrons in an energy level are distribuited according to the period in the periodic table they are.

So, if we have an atom in period 1, like Hydrogen (H), that atom would only have 1 level energy (n = 1) and in that level, we only have the sub level 1s.

Electrons in the 3d sublevel, are found mostly in all the transition metals of period 3, and it can go from 1 to 10 electrons. To be with the 3d sub level it's neccesary that the energy level to be 3.

energy levels beyond that, like n = 4, we have electrons occupying the 3d sub level, so, primordly, the 3d is found only in energy level 3.

Hope this helps

4 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

Use the table on the right to calculate each required quantity.

k0ka [10]

a. 35.8 KJ

b. 871 g

c. Fe

..........................................................................................

3 0

3 years ago