Answer:
λ = 0.0167 m = 16.7 mm
Explanation:
The wavelength of these radio waves can be found out by using the formula for the speed of radio waves:
v = fλ
where,
v = speed of radio waves = speed of light = 3 x 10⁸ m/s
f = frequency of radio waves = 18 GHz = 18 x 10⁹ Hz
λ = Wavelength = ?
Therefore,
3 x 10⁸ m/s = (18 x 10⁹ Hz)λ
λ = (3 x 10⁸ m/s)/(18 x 10⁹ Hz)
<u>λ = 0.0167 m = 16.7 mm</u>
C=0.10 mol/l
pH=-lg[H⁺]
HCl = H⁺ + Cl⁻
pH=-lgc
pH=-lg0.10=1.0
pH=1.0
Answer is: volume will be 6,7 L.
Boyle's Law: the pressure volume law - <span> volume of a given amount of gas held varies inversely with the applied pressure when the temperature and mass are constant.
p</span>₁V₁ = p₂V₂.
90 kPa · 5 L = 67 kPa · V₂.
V₂ = 90 kPa · 5 L / 67 kPa.
V₂ = 6,7 L, but same amount of oxygen.
Its C
a catalyst speeds up a reaction by offering the reaction an alternative reaction pathway with a lower activation energy
hope that helps
Answer:
8.8g of Al are necessaries
Explanation:
Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.
To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:
<em>Moles H2:</em>
PV = nRT; PV/RT = n
<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>
Replacing:
1atm*11L/0.082atmL/molK*273.15K = n
n = 0.491 moles of H2 must be produced
<em />
<em>Moles Al:</em>
0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required
<em />
<em>Mass Al -Molar mass: 26.98g/mol-:</em>
0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries
