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Tcecarenko [31]
3 years ago
15

URGENT! Will give 20 points.

Chemistry
2 answers:
Nataly [62]3 years ago
7 0
Idk why but for some reason the picture isnt ahowing up
Katena32 [7]3 years ago
6 0

Answer:

there is no picture...

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PH is 7.45. Calculate value of [H3O+] and [OH-]
lana66690 [7]

Answer:

The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M

Explanation:

We use the formulas:

pH= - log(H30+)  and Kwater=(H30+)x(OH-)

pH= - log(H30+)  ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M

Kwater=(H30+)x(OH-)

(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7

8 0
3 years ago
The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this p
earnstyle [38]

Answer:

The frequency is  f =  2,01 * 10^{15} \  Hz

Explanation:

From the question we are told that

   The energy required to ionize boron is E_b  =  801 KJ/mol

Generally the ionization energy of boron pre atom is mathematically represented as

     E_a  =  \frac{E_b}{N_A}

Here  N_A is the Avogadro's constant with value N_A  =  6.022*10^{23}

So

      E_a  =  \frac{801}{6.022*10^{23}}

=>     E_a  =  1.330 *10^{-18} \  J/atom

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

       E =  hf  =  E_a

=>     hf  =  E_a

Here h is the Planks constant with value h = 6.626 *10^{-34}

So

       f =  \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}

=>      f =  2,01 * 10^{15} \  Hz

8 0
3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
A 2. 75-l container filled with co2 gas at 25°c and 225 kpa pressure springs a leak. When the container is re-sealed, the pressu
Serjik [45]

The number of moles of gas lost is  0.0213 mol. It can be solved with the help of Ideal gas law.

<h3>What is Ideal law ?</h3>

According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.

PV = nRT.

Where,

  • p = pressure
  • V = volume (1.75 L = 1.75 x 10⁻³ m³)
  • T =  absolute temperature
  • n = number of moles
  • R =  gas constant, 8.314 J*(mol-K)

Therefore, the number of moles is

n = PV / RT

State 1 :

  • T₁ = (25⁰ C = 25+273 = 298 K)
  • p₁ = 225 kPa = 225 x 10³ N/m²

State 2 :

  • T₂ = 10 C = 283 K
  • p₂ = 185 kPa = 185 x 10³ N/m²

The loss in moles of gas from state 1 to state 2 is

Δn = V/R (P₁/T₁ - P₂/T₂ )

V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N

p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)

p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)

Therefore,

Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))

    = 0.0213 mol

Hence, The number of moles of gas lost is 0.0213 mol.

Learn more about ideal gas here ;

https://brainly.in/question/641453

#SPJ1

3 0
2 years ago
What instrument us normally used to measure atmospheric pressure
Anika [276]
The answer is Barometer.
3 0
3 years ago
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