Answer:
Cooking with open, exposed flame to prevent build up of CO₂ which will extinguish the fire
Explanation:
In the cooking process whereby food is heated by the combustion of cooking gas as follows;
C₄H₁₀ (g) + (13/2)O₂ → 4CO₂(g) + 5H₂O(g), ΔH = -2658 kJ·mol⁻¹
The product of the combustion must be allowed to escape freely to allow for more reactant molecules of oxygen and butane to effectively combine.
Whereby, the CO₂ is allowed to remain and accumulate at the reaction site, the CO₂ displaces the oxygen and reduces the butane such that as the CO₂ builds up without removal, the CO₂ and butane will left in the mixture while the fire is extinguished.
ANSWER
<em>2Ba + 2HBr → 2BaBr + H2</em>
<em>2BiCl3 + 3H2S → Bi2S3 + 6HCl</em>
<em>Br2 + 2KI → I2 + 2KBr</em>
<em>4Fe + 3O2 → 2Fe2O3</em>
<h3>
I HOPE THIS WILL HELP YOU IF NOT THEN SORRY HAVE A GREAT DAY:)</h3>
Answer:
Scale Factor: 1 : 2
X = 11
Explanation:
3 / 3 = 1
6 / 3 = 2
The scale factor should be 1:2
So 'X' should be 11 since 5.5 * 2 = 11
There are 3 significant figures
We can use two equations to solve this.
(1) - E = hf
E = Energy (J)
h = plank's constant (6.63 × 10⁻³⁴ J s)
f = frequency (Hz)
(2) - v = fλ
v = velocity of the wave (m/s)
f = frequency (Hz)
λ = wavelength (m)
the λ for the given wave is 625nm (625 x 10⁻⁹ m) and the velocity of the wave is equal to speed of light (3 x 10⁸ m/s) since the wave is a light.
hence we can find the frequency of the wave by using (2) formula.
3 x 10⁸ m/s = f x 625 x 10⁻⁹ m
f = 4.8 x 10¹⁴ Hz
by using calculated frequency and (1) formula, we can find the energy of the wave.
E = 6.63 × 10⁻³⁴ J s x 4.8 x 10¹⁴ Hz
E = 3.18 x 10⁻¹⁹ J
The energy of wave is less than energy needed to eject an electron.
Hence, the metal does not eject electron.
