Answer:
mass of water is 7.43 g
Explanation:
q = mc∆T
qsteam = 100 g × 1.864 J/(g °C) × (110 – 100) °C = 1864 J
qwater = mwater× 4.184 J/(g °C) × (95 – 35) °C = mwater× 251.04 J/g = qsteam = 1864 J
mwater = 1864 J / 251.04 J/g = 7.43 g
The color, shape, volume, mass, etc.
<span>Reaction of ethane combustion:
2C2H6 (g) + 7O2 (g) ----> 4CO2 (g) + 6H2O
According to the reaction, we can see that </span>C2H6 and CO2 have following stoichiometric ratio:
n(C2H6) : n(CO2) = 2 : 4
If we know the number of moles of ethane we can calculate moles of carbon dioxide:
2 x n(CO2) = 4 x n(C2H6)
n(CO2) = 2 x n(C2H6)
n(CO2) = 2 x 5.4 = 10.8 mole of CO2 <span>are produced</span>
Answer:
7.52 x 10²³ molecules N
Explanation:
multiply 1.25 mol of N2 by Avogadro's Number
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.
