Oxygen can be produced in the laboratory by the reaction

In the laboratory you dissolve 22.0 g of nickel(II) nitrate in a volumetric flask and add water to a total volume of 125

Ahat [919]

Answer:

1.364 M.

Explanation:

Molarity formula: M= n/v, where n is moles of solute, and v is liters of solution.

Now, we need to convert the grams of nickel to moles and the volume of water to liters.

125mL/1000= 0.125 L.

To convert nickel grams to moles, we need to take a look at it's chemical formula, which is:

$Ni(NO_{3} )_{2}$

Now we count how many molecules of each element we have:

Ni= 1

N= 2

O= 6

Calculate the weight (g) of each element (the values of g/mol can be found on the periodic table and they may vary slightly between one table and the other):

Ni: (1) (58.6934)= 58.6934

N= (2) (14.007)= 28.014

O= (6) (15.999)= 95.994

Sum all the values to obtain the total weight of 1 mole of this compound:

58.6934+28.014+95.994= 129(g/mole)

Now that we know the that 129 grams equal 1 mole of nickel(II) nitrate, we can convert the 22.0 g to moles:

129g ------- 1 mole

22.0g ----- x

x= (22*1)/129= 0.1705 moles.

Now, we have all the values needed to calculate the molarity of this solution. All we have to do is substitute the values in the formula:

M= (0.1705 moles) / (0.125 L)= 1.364 M.

3 0

2 years ago

1. If you have a sample of gas at a pressure of 16 atm, what will the pressure be if the volume is halved?

vagabundo [1.1K]

Answer:

1. The pressure will be 32 atm, twice the initial pressure.

2. The pressure will be 1.83 atm, one third of the initial pressure.

Explanation:

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.

This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1=P2*V2

1. In this case:

P1= 16 atm
V1
P2= ?
V2= V1÷2= $\frac{V1}{2}$ because the volume is halved.

So:

16 atm*V1= P2* $\frac{V1}{2}$

Solving:

$\frac{16 atm*V1*2}{V1}$ =P2

16 atm*2= P2

32 atm= P2

The pressure will be 32 atm, twice the initial pressure.

2. Now

P1= 5.5 atm
V1
P2= ?
V2= V1*3 because the volume is tripled.

So:

5.5 atm*V1= P2* V1*3

Solving:

$\frac{5.5 atm*V1}{3*V1}$ =P2

$\frac{5.5 atm}{3}$ = P2

1.83 atm= P2

The pressure will be 1.83 atm, one third of the initial pressure.

7 0

3 years ago

The combustion of 1.685 g of propanol (C3H7OH) increases the temperature of a bomb calorimeter from 298.00 K to 302.16 K. The he

AfilCa [17]

Answer:

ΔH = - 2020.57 kJ/mol

Explanation:

Given that :

mass of propanol = 1.685 g

the molar molar mass = 60 g/mol

Thus; the number of moles = mass/molar mass

= 1.685 g/60 g/mol

= 0.028 g/mol

However ;

ΔH = heat capacity C × Δ T

Given that:

The temperature increases from 298.00 K to 302.16 K.

Then ;

Δ T = 302.16 K - 298.00 K

Δ T = 4.16 K

heat capacity C = 13.60 kJ/K

∴

ΔH = 13.60 kJ/K × 4.16 K

ΔH = 56.576 kJ

The equation of the given reaction can be represented as :

$C_3H_7OH_{(l)}+\dfrac{3}{2}O_{2(g)} \to 3CO_{2(g)} +4H_2O_{(l)}$

Thus for 0.028 mol of heat liberated; ΔH = 56.576 kJ

For 1 mole of heat liberated now:

ΔH = 56.576 kJ/0.028 mol

ΔH = 2020.57 kJ/mol

SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;

ΔH = - 2020.57 kJ/mol

5 0

3 years ago

Read 2 more answers

The photograph shows one of the boats in a science club's toy boat race. All the boats were powered by identical

____ [38]

Answer:

Boat p

Explanation:

4 0

3 years ago

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you are investigating an organic compound. you discover the following: it contains no hydrogen atoms. what property do you expec

pashok25 [27]

This can be, for example, halogensubstituted hydrocarbons.
CCl₄, C₂F₆.

Or halides halocarboxylic acids, and other compounds.
O
II
Cl₃C-Cl

6 0

3 years ago

Read 2 more answers