Answer: 351
Explanation: From °C to kelvin we do
K=273+°C
From Kelvin to °C we do
°C=K-273 or -273+ K
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Answer:
Thx Have a Fantastic day :)
Explanation:
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
The de Broglie wavelength of a subatomic particle is 2.09 nm.
λ = h m v = h
momentum : wherein 'h' is the Plank's steady. This equation pertaining to the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated the use of this relation is de Broglie wavelength.
Frequency is the ratio of velocity and wavelength in relation to hurry. In evaluation, wavelength refers back to the ratio of velocity and frequency.
Wavelength is the gap between the crests of waves or a person's fashionable mind-set. An instance of wavelength is the gap between the crest of two waves. An instance of wavelength is while you and some other character share the equal standard attitude and might for that reason speak properly.
calculation is given in the image below
de Broglie wavelength λ = h/mv
= (6.626 * 10^-34)/9.1 * 10^-31 *351 *10^3
= 2.07 *10^-9
Hence, = 2.op nm
Learn more about de Broglie wavelength here:-brainly.com/question/16595523
#SPJ4