1.34 L of HF
Explanation:
We have the following chemical reaction:
Sn (s) + 2 HF (g) → SnF₂ (s) + H₂ (g)
First we calculate the number of moles of SnF₂:
number of moles = mass / molecular weight
number of moles of SnF₂ = 5 / 157 = 0.03 moles
From the chemical reaction we see that 1 mole of SnF₂ are produced from 2 moles of SnF₂. This will mean that 0.03 moles of SnF₂ are produced from 0.06 moles of HF.
Now at standard temperature and pressure (STP) we can use the following formula to calculate the volume of HF:
number of moles = volume / 22.4 (L/mole)
volume of HF = number of moles × 22.4
volume of HF = 0.06 × 22.4 = 1.34 L
Learn more about:
problems with gases at STP
brainly.com/question/8857334
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1) The trails left by an electron as it moves around the nucleus
The electron model dictates that the electrons have no fixed position so it traces their path.
2) 8
Atomic number is equivalent to proton number
3) Its mass is lowered, but it is still the same element.
The element's identity is due to the number of protons; however, neutrons play a large role in an atom's mass. Thus, the mass will decrease but the element will be the same. Such variants are called isotopes.
The reaction is
CaC₂(s) + 2H₂O (l) -----> Ca(OH)₂ (s) + C₂H₂ (g)
As we have data of gas ethyne (or acetylene), C₂H₂
We can calculate the moles of acetylene and from this we can estimate the mass of calcium carbide taken
the moles of acetylene will be calculated using ideal gas equation
PV =nRT
R = gas constant = 0.0821 Latm/molK
T = 385 K
V = volume = 550 L
P = Pressure = 1.25 atm
n = moles = ?
n = PV /RT = 1.25 X 550 / 0.0821 X 385 = 21.75 mol
As per balanced equation these moles of acetylene will be obtained from same moles of calcium carbide
moles of calcium carbide = 21.75mol
molar mass of CaC₂ = 40 + 24 = 64
mass of CaC₂ = moles X molar mass = 21.75 X 64 = 1392g
The answer is b
Explanation:
Element X is sodium.
If there is one more shell of electrons, then this means it is one period below.
Since there is one less valence electron, it is one group to the left.
