All categories

3 years ago

Given the unbalanced equation:

Chemistry

1 answer:

maw [93]3 years ago

8 0

Answer:

Option (2) 2

Explanation:

NO3- + 4H+ + Pb → Pb2+ + NO2 + 2H2O

The equation above can be balance as follow:

There are 3 atoms of the left side and a total of 4 atoms on the right side. It can be balance by putting 2 in front NO3- and 2 in front of NO2 as shown below:

2NO3- + 4H+ + Pb → Pb2+ + 2NO2 + 2H2O

Now the equation is balanced.

The coefficient of NO2 is 2

You might be interested in

1. Which groups of atoms become MORE stable when they lose an electron?

FinnZ [79.3K]

Probably Metal atoms?

8 0

3 years ago

Brass is a substitutional alloy consisting of a solution of copper and zinc. A particular sample of red brass consisting of 79.0

Bas_tet [7]

Answer: The molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

Explanation:

Solute is the substance which is present in smaller proportion in a mixture and solvent is the substance which is present in larger proportion in a mixture.

We are given:

(m/m) % of Cu = 79 %

This means that 79 g of copper is present in 100 grams of alloy.

(m/m) % of Zn = 21 %

This means that 21 g of zinc is present in 100 grams of alloy.

As, zinc is present in smaller proportion. So, it is solute and copper is the solvent.

Calculating molality of zinc:

To calculate molality of the zinc, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$m_{solute}$ = Given mass of solute (Zinc ) = 21 g

$M_{solute}$ = Molar mass of solute (Zinc) = 65.3 g/mol

$W_{solvent}$ = Mass of solvent (copper) = 79 g

Putting values in above equation, we get:

$\text{Molality of Zinc}=\frac{21\times 1000}{65.3\times 79}\\\\\text{Molality of zinc}=4.07m$

Calculating molarity of zinc:

To calculate volume of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = $8740kg/m^3=\frac{8740kg\frac{1000g}{1kg}}{1m^3\times \frac{10^6cm^3}{1m^3}}=\frac{8740g}{1000cm^3}=8.740g/cm^3$

(Conversion factors used are: 1 kg = 1000 g & $1m^3=10^6cm^3$ )

Mass of solution = 100 g

Putting values in above equation, we get:

$8.740g/cm^3=\frac{100.0g}{\text{Volume of zinc}}\\\\\text{Volume of zinc}=11.44cm^3$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molar mass of zinc = 65.3 g/mol

Volume of solution = $11.44cm^3=11.44mL$ (Conversion factor: $1cm^3=1mL$ )

Mass of zinc = 21.0 g

Putting values in above equation, we get:

$\text{Molarity of zinc}=\frac{21\times 1000}{65.3\times 11.44}=28.11M$

Hence, the molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

4 0

3 years ago

How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?

shepuryov [24]

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically, molarity = $\frac{no. of moles}{Volume (in L) of solution}$

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

molarity = $\frac{no. of moles}{Volume of solution in liter}$

0.0800 M = $\frac{no. of moles}{0.05 L}$

no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

No. of moles = $\frac{mass in grams}{molar mass}$

mass in grams = $no. of moles \times molar mass of CuSO_{4}.5H_{2}O$

= $1.6 mol \times 249.68 g/mol$

= 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

4 0

3 years ago

Can one atom in this type of reaction win the "tug of war"? What might happen if it did?

Luden [163]

Answer:

POLAR BONDS will form.

A polar bond is a covalent bond between two atoms where the electrons forming the bond are unequally distributed. This causes the molecule to have a slight electrical dipole moment where one end is slightly positive and the other is slightly negative.

7 0

3 years ago

How to do compare and contrast energies

julsineya [31]

Is it kinetic or potential energy Or frequency and wavelength

3 0

3 years ago