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3 years ago

A scoop of potassium oxide was placed in

Chemistry

1 answer:

kotykmax [81]3 years ago

8 0

Answer:

I just need points sorry

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Consider an atom having four distinct energy levels. If an electron is able to make transitions between any two levels, how many

Kryger [21]

Energy levels are the electron shells where electrons are found at a fixed distance from the nucleus of the atom. The atom could emit 6 different wavelengths.

<h3>What is wavelength?</h3>

A wavelength is a distance between the adjacent crests in wave signals propagated in a system. Wavelength $(\rm \lambda )$ is in inverse relation to the frequency of the wave.

When an electron jumps from energy level 1 to 2, 1 to 3, and 1 to 4 one wavelength each is present. Hence, making the total wavelength to be 3, in transition from the first energy level.

Similarly, from energy levels, 2 to 3 and 2 to 4, a total of 2 wavelengths, and from energy levels 3 to 4 one wavelength is produced.

So the total different wavelengths of the radiation that can be emitted will be 3 + 2 + 1 = 6.

Therefore, 6 different wavelengths of radiation will be emitted by the atom.

Learn more about wavelengths here:

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7 0

2 years ago

An unknown metal absorbs 61 J of heat, and its temperature increases by 29∘C. What is the heat capacity of the metal?

aniked [119]

Answer:

2.103 J/C

Explanation:

Quantity of heat = Heat Capacity * Temperature change

Heat Capacity = Quantity of heat / Temperature Change

Heat Capacity = 61/29

Heat Capacity = 2.103 J/C

3 0

3 years ago

The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni

ValentinkaMS [17]

Answer:

" $6.7\times 10^{-4} \ atm$ " is the right answer.

Explanation:

Given:

Partial pressure of $N_2$ ,

= 0.20 atm

Partial pressure of $H_2$ ,

= 0.15 atm

$K_p = 1.5\times 10^3$ at $400^{\circ} C$

As we know,

⇒ $K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}$

By putting the values, we get

$1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}$

$pNH_3^2 = \frac{0.000675}{1.5\times 10^3}$

$=6.7\times 10^{-4} \ atm$

3 0

3 years ago

It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis

Norma-Jean [14]

Answer:

The overall balanced combustion reaction is written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

$(F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = 23.562$

the higher heating values $(HHV)_f$ per unit mass of LPG = 49.9876 MJ/kg

the lower heating values $(LHV)_f$ per unit mass of LPG = 46.4933 MJ/kg

Explanation:

The stoichiometric equation can be expressed as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2 \ + \ bH_2O \ + \ cN_2$

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

$x = \frac{2a+b}{2} \\ \\ x = \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95$

Equating the coefficient of Nitrogen

$c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612$

Therefore, The overall balanced combustion reaction can now be written as :

$0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2$

Now; To determine the stoichiometric F/A and A/F ratios; we have:

$(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\ (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424$

$(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\ (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562$

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel $M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})$

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

$m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8} = 0.7 \\ \\ \\ m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06 \\ \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6} = 0.24$

Finally; calculating the higher heating values $(HHV)_f$ per unit mass of LPG; we have:

$(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg$

calculating the lower heating values $(LHV)_f$ per unit mass of LPG; we have:

$(LHV)_f = (HHV)_f - \delta H_w \\ \\ (LHV)_f = (HHV)_f - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f = 49.9876 \ MJ/kg - [\frac{3.8*18}{44.2}*2.258 \ MJ/kg] \\ \\ (LHV)_f = 46.4933 \ M/kg$

7 0

3 years ago

What is the wavelength of a 3 x 1012 Hz infrared wave?

Alenkasestr [34]

Answer:

Explanation:

wavelength = speed of light/ frequency

= (3 x 10^8 m/s) / (3 x 10^12 Hz)

= 1 x 10^-4 m

5 0

2 years ago