A chemical property of a substance is a certain characteristic that can only be observed by participating in a chemical reaction. Alternatively, a chemical property of a substance is something that can only be observed when the substance undergoes a chemical change.
I'm not sure what you're supposed to do with the first four boxes; all four are examples of chemical properties. Do you have to name the specific type of chemical property as given in the description? If so, the following would be my answers:
Flammability/Combustibility: The ability of a substance to burn.
The next two are quite strange; I'm not aware of a term that cleanly describes reactivity with water or acid. I suspect that, given the level of the material here, the general property of "reactivity" might be the answer for both the second and third descriptions
(Water-)reactivity: Some substances react when put in water.
(Acid-)reactivity: Some substances react when put in acid.
Light sensitivity: Light can interact with some things to form new substances.
As for the chart, I've filled it in as shown in the attached image. Please take care to double-check what I've written; in particular, when it comes to the property, I might have used a different term from what you were taught in class or provided in some other resource that I don't have access to. I've also color-coded qualitative/quantitative and physical/chemical for your convenience.
Answer is: <span>de Broglie wavelength of a proton is </span>3,4·10⁻⁵ nm.
v(proton) = 0,038 · 3·10⁸ m/s.
v(proton) = 1,14·10⁷ m/s; speed of proton.
m(proton) = 1,67·10⁻²⁷ kg.
h = 6,62607004·10⁻³⁴ m²·kg/s; Planck constant.
λ(proton) = h / m(proton) · v(proton).
λ(proton) = 6,62607004·10⁻³⁴ m²·kg/s ÷ (1,67·10⁻²⁷ kg · 1,14·10⁷ m/s).
λ(proton) = 3,48·10⁻¹⁴ m · 10⁹ nm/m = 3,4·10⁻⁵ nm.
Answer:
65.2L
Explanation:
Using the general gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (Litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (Kelvin)
According to the information provided in this question,
P = 1.631 atm
V = ?
n = 4.3 moles
T = 28°C = 28 + 273 = 301K
Using PV = nRT
V = nRT/P
V = 4.3 × 0.0821 × 301 ÷ 1.631
V = 106.26 ÷ 1.631
V = 65.15
Volume of the gas = 65.2L
Answer:
V = 22.42 L/mol
N₂ and H₂ Same molar Volume at STP
Explanation:
Data Given:
molar volume of N₂ at STP = 22.42 L/mol
Calculation of molar volume of N₂ at STP = ?
Comparison of molar volume of H₂ and N₂ = ?
Solution:
Molar Volume of Gas:
The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol
Molar volume can be calculated by using ideal gas formula
PV = nRT
Rearrange the equation for Volume
V = nRT / P . . . . . . . . . (1)
where
P = pressure
V = Volume
T= Temperature
n = Number of moles
R = ideal gas constant
Standard values
P = 1 atm
T = 273 K
n = 1 mole
R = 0.08206 L.atm / mol. K
Now put the value in formula (1) to calculate volume for 1 mole of N₂
V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm
V = 22.42 L/mol
Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.
Answer:
temperate, hot, degradation, California
Explanation:
