Answer:
A. There was still 140 ml of volume available for the reaction
Explanation:
According to Avogadro's law, we have that equal volumes of all gases contains equal number of molecules
According to the ideal gas law, we have;
The pressure exerted by a gas, P = n·R·T/V
Where;
n = The number of moles
T = The temperature of the gas
R = The universal gas constant
V = The volume of the gas
Therefore, given that the volumes and number of moles of the removed air and added HCl are the same, the pressure and therefore, the volume available for the reaction will remain the same
There will still be the same volume available for the reaction.
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.
Answer:
Chemical name for As2S3 is Arsenic sulfide
Explanation:
Answer:
It has been balanced by using the half-reaction method.
Explanation:
I- and IO3- gives I2
We divide the reaction into two half-reactions
(2 I- >> I2 + 2e-) x5 ( oxidation : I goes from -1 to 0 )
2 IO3- + 12H+ + 10e- >> I2 + 6H2O ( reduction : I goes from +5 to 0 )
10 I- >> 5I2 + 10e-
2IO3- + 12H+ + 10e- >> I2 + 6H2O
-----------------------------------------------------
10 I- + 2IO3- + 12H+ >> 6I2 + 6H2O
To get the smallest numbers we divide by 2 :
5 I- + IO3- + 6H+ >> 3I2 + 3H2O
M=43lb = 19,5kg
If 115mg --------- is for --------- 1kg
so
x ---------- is for --------- 19,5kg
x = 19,5kg * 115mg / 1kg
x = 2242,5 mg
