These are two questions and two answers
Question 1.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 9.11 × 10⁻³¹ kg
b) λ = 3.31 × 10⁻¹⁰ m
c) c = 3.00 10⁸ m/s
d) s = ?
<u>2) Formula:</u>
The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Solve for s:
Substitute:
- s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg × 3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s
To express the speed relative to the speed of light, divide by c = 3.00 10⁸ m/s
- s = 2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³
Answer: s = 7.33 × 10 ⁻³ c
Question 2.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 45.9 g (0.0459 kg)
b) s = 70.0 m/s
b) λ = ?
<u>2) Formula:</u>
Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Substitute:
- λ = 6.626×10⁻³⁴J.s / ( 0.0459 kg × 70.0 m/s) = 2.06 × 10 ⁻³⁴ m
As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.
Answer: 2.06 × 10 ⁻³⁴ m.
1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr /0.0135= 1 mol Cr
0.0538 mol O/0.035= 4 mol Cr
K2CrO4
Answer:
Answer is: 20 min.Bismuth 214 decay mode to lead is beta minus decay and alpha decay.Beta decay is radioactive decay in which a beta ray and a neutrino are emitted from an atomic nucleus.
There are two types of beta decay: beta minus and beta plus. In beta minus decay, neutron is converted to a proton and an electron and an electron.Alpha decay is radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus) and transforms into an atom with an atomic number that is reduced by two and mass number that is reduced by four.
Explanation:
Answer:
hello your question lacks the required reaction pairs below are the missing pairs
Reaction system 1 :
A + B ⇒ D ![-r_{1A} = 10exp[-8000K/T]C_{A}C_{B}](https://tex.z-dn.net/?f=-r_%7B1A%7D%20%20%3D%2010exp%5B-8000K%2FT%5DC_%7BA%7DC_%7BB%7D)
A + B ⇒ U 
Reaction system 2
A + B ⇒ D 
B + D ⇒ U 
Answer : reaction 1 : description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1
condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2
reaction 2 :
description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1
condition to maximize selectivity:
to increase selectivity the concentration of D should be minimal
Explanation:
Reaction system 1 :
A + B ⇒ D ![-r_{1A} = 10exp[-8000K/T]C_{A}C_{B}](https://tex.z-dn.net/?f=-r_%7B1A%7D%20%20%3D%2010exp%5B-8000K%2FT%5DC_%7BA%7DC_%7BB%7D)
A + B ⇒ U 
the selectivity of D is represented using the relationship below
hence SDu = 1/10 * 
description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1
condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2
Reaction system 2
A + B ⇒ D 
B + D ⇒ U 
selectivity of D

hence Sdu = 
description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1
condition to maximize selectivity:
to increase selectivity the concentration of D should be minimal