4. Chlorine exists in two forms

You are given 50.0 ml of 1.50 m hno3. how many ml of 0.81 m naoh are needed to neutralize it?

svet-max [94.6K]

V(HNO₃) = 50.0 mL in liters = 50.0 /1000 =0.05 L
M(HNO₃) = 1.50 M

Number of moles HNO₃ :

n = M x V

n = 1.50 x 0.05

n = 0.075 moles of HNO₃

HNO₃ + NaOH = H₂O + NaNO₃

1 mole HNO₃ -------- ---1 mole NaOH
0.075 moles HNO₃ ---- ?

moles NaOH = 0.075 * 1 / 1

= 0.075 moles of NaOH

V ( NaOH ) :

M = n / V

0.81 = 0.075 / V

V = 0.075 / 0.81

V = 0.0925 L or 92.5 mL 

hope this helps!

7 0

3 years ago

The SI unit of time is the second, which is defined as 9,192,631,770 cycles of radiation associated with a certain emission proc

elixir [45]

For waves:
speed = frequency x wavelength
speed = 3 x 10⁸
frequency = 9192631770 Hz
wavelength = 3 x 10⁸ / (9192631770)
wavelength = 0.033 meters
This wavelength lies in the region of microwaves.

4 0

3 years ago

Read 2 more answers

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) Minimum mass of air

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

3 years ago

Which of the following shows an accurate combustion reaction?

Jet001 [13]

Answer:

Cu + O₂ → CuO₂

Explanation:

A combustion reaction is the reaction with oxygen along with the release of energy in form of heat or light.

Organic compounds (like CH₄) undergo combustion forming water and CO₂.

The combustion reaction of CH₄ is:

CH₄ + 2O₂ → CO₂ + 2H₂O

Hence, the first equation from the choices is not showing the combustion reaction of CH₄.

Not only organic compounds can undergo combustion. Metals and no metals can undergo combustion, i.e. metals and no metals can react with oxygen releasing light or heat.

The reaction of copper and oxygen (second choice) is a combustion reaction:

Cu + O₂ → CuO₂

The formation of water (2H₂ + O₂ → 2H₂O) is other example of a combustion reaction where no organic compounds are involved.

On the other hand, the other two equations from the choice list are not reactions with oxygen, so they do not show combustion reactions.

5 0

4 years ago

Examine the statement.

sweet [91]

This is an exothermic

8 0

3 years ago

4. Chlorine exists in two forms - chlorine-35 and